I would like to point out that even though the OP calls the property in question expansivity the kind of expansivity he is talking about is stronger than the kind of expansivity the other two answers seem to focus on. As such, there is an elementary proof (that is, the proof does not require any technology apart from basic metric space theory and some combinatorics). Still, it is slightly more involved than the argument the OP presents, as Moishe Kohan pointed out said argument is not quite sufficient.
First let me give some definitions. Let $(X,d)$ be a compact metric space, $\epsilon^\ast\in\mathbb{R}_{>0}$, and $\mathcal{F}\subseteq C^0(X,X)$ be a collection of continuous self-maps of it. $\mathcal{F}$ is called $\epsilon^\ast$-expansive if
$$\forall x,y\in X: \sup_{f\in\mathcal{F}} d(f(x),f(y))\leq \epsilon^\ast \implies x=y.$$
Note that if $\mathcal{G}\subseteq C^0(X,X)$ is another family of continuous self-maps and $\mathcal{F}\subseteq \mathcal{G}$, then the $\epsilon^\ast$-expansivity of $\mathcal{F}$ would imply the $\epsilon^\ast$-expansivity of $\mathcal{G}$, so that expansivity gets stronger if the family is made smaller.
If $f\in C^0(X,X)$, then $f$ is called one-sided $\epsilon^\ast$-expansive (alternatively: positively $\epsilon^\ast$-expansive or forward $\epsilon^\ast$-expansive) if the family $\{f^n\,|\, n\in\mathbb{Z}_{\geq0}\}$ is $\epsilon^\ast$-expansive.
Similarly if $f\in \operatorname{Homeo}(X)$, then $f$ is called one-sided $\epsilon^\ast$-expansive if the family $\{f^n\,|\, n\in\mathbb{Z}_{\geq0}\}$ is $\epsilon^\ast$-expansive. In the case of homeomorphisms one can also consider the backward iterates, and $f$ is called two-sided $\epsilon^\ast$-expansive (or simply $\epsilon^\ast$-expansive, for reasons that will be clear in a bit) if the family $\{f^n\,|\, n\in\mathbb{Z}\}$ is $\epsilon^\ast$-expansive. Since the smaller the family the stronger the expansivity, one-sided expansivity of a homeomorphism is stronger than two-sided expansivity.
Any one of the above objects is called expansive if it's $\epsilon^\ast$-expansive for some $\epsilon^\ast\in\mathbb{R}_{>0}$.
For $f$ a homeomorphism of the compact metric space $(X,d)$, one can introduce new distances on $X$ (In ergodic theory circles these are often called Bowen metrics nowadays). I will be using them because to me they make the quantifier logistics easier to handle. Namely, for any subset $S\subseteq \mathbb{R}$ let us put
$$d_S=d_S^f:X\times X\to \mathbb{R}_{\geq0},\quad (x,y)\mapsto \sup_{n\in S\cap \mathbb{Z}} d(f^n(x),f^n(y)).$$
Thus e.g. $d_{[0,n]}^f(x,y)$ gives the largest distance that can be achieved between the orbits of $x$ and $y$ when $f$ acts on them for $n$ time units (the points in the respective orbits ought to be synchronized in time, so that we only compare $f^k(x)$ and $f^k(y)$).
Using these new metrics note that one-sided $\epsilon^\ast$-expansivity of $f$ is equivalent to:
$$\forall x,y\in X: d_{\mathbb{Z}_{\geq0}}^f(x,y)\leq\epsilon^\ast\implies x=y.$$
There is an alternative characterization of $\epsilon^\ast$-expansivity (I won't prove it, but the ideas in the next section are adaptations of the proof of it.):
Prop.: Let $(X,d)$ be compact metric, $\epsilon^\ast\in\mathbb{R}_{>0}$.
- Let $f:X\to X$ be continuous. Then $f$ is one-sided $\epsilon^\ast$-expansive iff
$$\forall \epsilon\in\mathbb{R}_{>0},\exists M\in\mathbb{Z}_{>0},\forall x,y\in X: d_{[0,M]}^f(x,y)\leq \epsilon^\ast \implies d(x,y)<\epsilon.$$
- Let $f: X\to X$ be a homeomorphism. Then $f$ is two-sided $\epsilon^\ast$-expansive iff
$$\forall \epsilon\in\mathbb{R}_{>0},\exists M\in\mathbb{Z}_{>0},\forall x,y\in X: d_{[-M,M]}^f(x,y)\leq \epsilon^\ast \implies d(x,y)<\epsilon.$$
We claim that no infinite compact metric space can carry a one-sided expansive homeomorphism. According to Gottschalk & Hedlund's Topological Dynamics (pp.85-86, Thm.10.30) this was first proved by Schwartzman in his 1952 thesis. Over the years it seems this result has been optimized; I'll be following Coven & Keane's paper "Every compact metric space that supports a positively expansive homeomorphism is finite". Also see Aoki & Hiraide's Topological Theory of Dynamical Systems - Recent Advances (p.45, Thm.2.2.12) for a more high-brow proof.
The proof uses the following lemma. Roughly speaking it says that in the case of an injective one-sided $\epsilon^\ast$-expansive map, if the orbits of any two points are $\epsilon^\ast$-indistinguishable for a sufficiently long time duration in the future, then any two such indistinguishable points will be $\epsilon^\ast$-indistinguishable also before said duration in the future.
Lemma: Let $(X,d)$ be compact metric, $\epsilon^\ast\in\mathbb{R}_{>0}$, $f:X\to X$ be continuous. If $f$ is injective and one-sided $\epsilon^\ast$-expansive, then
$$\exists M^\ast\in\mathbb{Z}_{>0}, \forall x,y\in X, \forall p\in\mathbb{Z}_{\geq0}: d_{[p+1,p+M^\ast]}^f(x,y)\leq \epsilon^\ast \implies d_{[0,p]}^f(x,y)\leq \epsilon^\ast$$
Pf.: First let us prove the statement for $p=0$, that is, we claim first that
$$\exists M^\ast\in\mathbb{Z}_{>0}, \forall x,y\in X: d_{[1,M^\ast]}^f(x,y)\leq \epsilon^\ast \implies d(x,y)\leq \epsilon^\ast.$$
Suppose otherwise. Then there are two sequences $x_\bullet,y_\bullet:\mathbb{Z}_{\geq0}\to X$ such that
$$\forall n\in\mathbb{Z}_{\geq0}: d_{[1,n]}^f(x_n,y_n)\leq \epsilon^\ast \quad\text{ and }\quad d(x_n,y_n)> \epsilon^\ast.$$
By compactness up to subsequences we have $\lim_{n\to\infty} x_n=x^\ast\in X$, $\lim_{n\to\infty} y_n=y^\ast\in X$, and since the terms are at least $\epsilon^\ast$ apart we have that $x^\ast\neq y^\ast$. Let $k\in\mathbb{Z}_{\geq1}$. Then for $n\in\mathbb{Z}_{\geq k}$: $d(f^k(x_n),f^k(y_n))\leq d_{[1,n]}^f(x_n,y_n)\leq\epsilon^\ast$, thus taking the limit as $n\to\infty$ we have $d(f^k(x^\ast), f^k(y^\ast))\leq \epsilon^\ast$. In other words
$$d_{\mathbb{Z}_{\geq1}}^f(x^\ast,y^\ast)=d_{\mathbb{Z}_{\geq0}}^f(f(x^\ast),f(y^\ast))\leq \epsilon^\ast.$$
By the one-sided $\epsilon^\ast$-expansivity of $f$, $f(x^\ast)=f(y^\ast)$, but $f$ is assumed to be injective, a contradiction. Thus there is an $M^\ast$ as claimed.
Next we claim that this $M^\ast$ works uniformly for $x,y$ and $p$ as in the statement. Suppose $p\in\mathbb{Z}_{>0}$ and fix $x,y\in X$ with $d_{[p+1, p+M^\ast]}^f(x,y)=d_{[1,M^\ast]}^f(f^p(x),f^p(y))\leq \epsilon^\ast$. Applying the argument for the $p=0$ case to $f^p(x)$ and $f^p(y)$, we get $d(f^p(x),f^p(y))\leq\epsilon^\ast$, whence we also have
$$d(f^p(x),f^p(y))\leq d_{[p,(p-1)+M^\ast]}^f(x,y)\leq d_{[p, p+M^\ast]}^f(x,y)\leq \epsilon^\ast.$$
Again, by the argument for the $p=0$ case applied to $f^{p-1}(x)$ and $f^{p-1}(y)$ we get
$$d(f^{p-1}(x),f^{p-1}(y))\leq d_{[p-1,(p-2)+M^\ast]}^f(x,y)\leq d_{[p-1, (p-1)+M^\ast]}^f(x,y)\leq \epsilon^\ast.$$
Iterating this process we have the result.
Theorem: Let $(X,d)$ be compact metric, $f:X\to X$ be continuous. If $f$ is injective and one-sided expansive, then $X$ is finite.
Pf.: Say $f$ is $\epsilon^\ast$-expansive for $\epsilon^\ast\in\mathbb{R}_{>0}$ and take $M^\ast\in\mathbb{Z}_{>0}$ be as in the previous lemma. For $z\in X$, put
$$U_z:=\left\{y\in X\,\left|\, d_{[1,M^\ast]}^f(z,y)<\dfrac{\epsilon^\ast}{2}\right.\right\}.$$
By compactness for some finite set $\{z_1,...,z_r\}\subseteq X$, $X=\bigcup_{s=1}^r U_{z_s}$. We claim that $\#(X)\leq r$. Suppose otherwise and fix a subset $S\subseteq X$ with $\#(S)=r+1$. For $k\in\mathbb{Z}_{\geq0}$, $f^k(S)$ has exactly $r+1$ elements, thus $S$ has at least two distinct elements that are mapped to the same $U_{z_s}$ by $f^k$. This way we may choose two sequences $x_\bullet,y_\bullet:\mathbb{Z}_{\geq0}\to S$ such that
$$\forall k\in\mathbb{Z}_{\geq0}: x_k\neq y_k, \,\,d_{[1,M^\ast]}^f(f^k(x_k),f^k(y_k))=d_{[k+1,k+M^\ast]}^f(x_k,y_k)\leq \epsilon^\ast.$$
By compactness up to subsequences we have $\lim_{n\to\infty} x_n=x^\ast\in S$, $\lim_{n\to\infty} y_n=y^\ast\in S$, and since $S$ is finite for some $K\in\mathbb{Z}_{\geq0}$, $x_k=x^\ast$ and $y_k=y^\ast$ for any $k\in\mathbb{Z}_{\geq K}$. In particular $x^\ast\neq y^\ast$ and $d_{[k+1,k+M^\ast]}^f(x^\ast,y^\ast)\leq \epsilon^\ast$ for any $k\in\mathbb{Z}_{\geq K}$. Applying the lemma with $p=K$ we also have $d_{[0,K]}^f(x^\ast,y^\ast)\leq \epsilon^\ast$, so that $d_{\mathbb{Z}_{\geq0}}^f(x^\ast,y^\ast)\leq \epsilon^\ast$, a contradiction to the $\epsilon^\ast$-expansivity of $f$.
It's clear that the theorem implies that the circle does not carry any one-sided expansive homeomorphism.