Such subsets of the line do exist. This explicitly follows from the proof of Theorem $2$ of Brian M. Scott, On the existence of totally inhomogeneous spaces [PDF], Proc. Amer. Math. Soc. $\mathbf{51}$ $(1975)$, pp. $489$-$493$, since the space $[0,1]$ with the usual topology satisfies the hypotheses of that theorem. Indeed, the proof shows that there is a pairwise disjoint, pairwise non-homeomorphic family of $\mathfrak{c}=2^\omega$ such sets, each of which is dense in $[0,1]$. They are constructed (very non-constructively!) by transfinite recursion.
Added: In the case $X=[0,1]$ the argument in the paper actually produces a family $\{R_\xi:\xi<2^\omega\}$ of pairwise disjoint, dense subspaces of $X$ such that if $\xi\le\eta<2^\omega$, $U$ is a non-empty open subset of $R_\xi$, and $V$ is a non-empty open subset of $R_\eta$, then $U$ and $V$ are not homeomorphic. Here I’ll simplify the argument slightly to produce a single such set.
Let $\mathscr{C}$ be the set of continuous partial functions from $X$ to $X$. Say the $f\in\mathscr{C}$ is a displacement if there is an $A\subseteq\operatorname{dom}f$ of cardinality $2^\omega$ such that $A\cap f[A]=\varnothing$. Let
$$\mathscr{C}_0=\{f\in\mathscr{C}:f\text{ is a displacement and }\operatorname{dom}f\text{ is a }G_\delta\}\;;$$
then $|\mathscr{C}_0|=2^\omega$, since $X$ has only $2^\omega$ $G_\delta$ subsets and is hereditarily separable, so we can index $\mathscr{C}_0=\{f_\xi:\xi<2^\omega\}$. By a standard result (e.g., Theorem $\mathbf{4.3.20}$ in Engelking’s General Topology) each displacement in $\mathscr{C}$ can be extended to an element of $\mathscr{C}_0$.
Let $\mathscr{K}$ be the family of compact subsets of $X$ of cardinality $2^\omega$; $|\mathscr{K}|=2^\omega$, so we can let $\mathscr{K}=\{K_\xi:\xi<2^\omega\}$ be an enumeration of $\mathscr{K}$. Let $D=\{d_\xi:\xi<2^\omega\}$ be an enumeration of $2^\omega\times2^\omega$, and let $\tau=\{U_\xi:\xi<2^\omega\}$ enumerate the topology $\tau$ of $X$.
Suppose that $\eta<2^\omega$, and points $p_\xi,q_\xi,z_\xi\in X$ have been chosen for $\xi<\eta$, and let $d_\eta=\langle\alpha,\beta\rangle$. Let $A_\eta=\{p_\xi:\xi<\eta\}\cup\{q_\xi:\xi<\eta\}\cup\{z_\xi:\xi<\eta\}$. Choose $p_\eta\in(\operatorname{dom}f_\alpha)\setminus A_\eta$ such that $f_\alpha(p_\eta)\ne p_\eta$, and if possible such that $p_\eta\in U_\beta$. Let $q_\eta=f_\alpha(p_\eta)$, and let $z_\eta\in K_\alpha\setminus\{p_\eta,q_\eta\}$. This is always possible, since $f_\alpha\in\mathscr{C}_0$, and $|K_\alpha|=2^\omega$. Let $R=\{p_\xi:\xi<2^\omega\}\cup\{z_\xi:\xi<2^\omega\}$; clearly $|R_\eta|=2^\omega$.
Suppose that $V,W\in\tau$, $\varnothing\ne V\cap R\ne W\cap R\ne\varnothing$; we want to show that $V\cap R$ and $W\cap R$ are not homeomorphic, so suppose that $h:V\cap R\to W\cap R$ is a homeomorphism. Replacing $h$ by $h^{-1}$ if necessary, we may assume that $(V\cap R)\setminus(W\cap R)\ne\varnothing$. Fix $x\in(V\cap R)\setminus(W\cap R)$; then $h(x)\ne x$, so there are disjoint $U,G\in\tau$ such that $x\in U\subseteq V$, $h(x)\in G\subseteq W$, and $h[U\cap R]=G\cap R$. $|U\cap R|=2^\omega$, so $h$ is a displacement, and there is an $\alpha<2^\omega$ such that $h=f_\alpha\upharpoonright(V\cap R)$. $U=U_\beta$ for some $\beta<2^\omega$, and there is a $\xi<2^\omega$ such that $d_\xi=\langle\alpha,\beta\rangle$. But then $p_\xi\in U_\beta\cap R\subseteq V\cap R$, so $h(p_\xi)\in W\cap R$, but $h(p_\xi)=f_\alpha(p_\xi)=q_\xi\notin R$. This contradiction shows that no such homeomorphism $h$ can exist.
