Let $C\subset {\mathbb C}$ be a connected, locally connected subset of the Euclidean plane separating $0$ from $\infty$ in the Riemann sphere $S^2= {\mathbb C}\cup\{\infty\}$. (WLOG we can work with these points instead of your points $x, y$.)
Then $S^2\setminus C$ consists of at least two components; all complementary components are simply connected. Let $U$ denote the connected component of $0$ in $S^2\setminus C$. Let $f: D\to U$ be a Riemann mapping from the open unit disk $D$ onto $U$ such that $f(0)=0$. Since $C$ is locally connected, the map $f$ extends continuously to the closed disk
$\bar{D}$ so that $K=f(\partial D)\subset C$; I retain the notation $f$ for the extension. (See my answer here.)
Our goal is to find a Jordan curve in $K$ separating $0$ from $\infty$. This would have been easy if $f$ were 1-1, of course, which is, in general, will not be true. Nevertheless, the restriction
of $f$ to $S^1=\partial D$ defines a loop $\sigma$ in ${\mathbb C}^*={\mathbb C}\setminus \{0\}$.
I leave the proof of this lemma to you:
Lemma 1. The loop $\sigma$ is not null-homotopic in ${\mathbb C}^*$; it actually represents a generator of $\pi_1({\mathbb C}^*)$ once
we orient $S^1$.
Suppose that $f$ restricted to $S^1=\partial D$ is not 1-1. Let $p, q\in \partial D$ be distinct points such that $f(p)=f(q)$. Then $p, q$ are end-points of two distinct arcs $\alpha_{pq}, \beta_{pq}$ in $S^1=\partial D$ oriented the same way as $S^1$. Restricting $f$ to these arcs we obtain loops $\gamma, \delta$ in $K$ based at the point $z=f(p)=f(q)$.
Lemma. Exactly one of the loops $\gamma, \delta$ is null-homotopic in ${\mathbb C}^*={\mathbb C}\setminus \{0\}$.
Proof. The product $[\gamma]\cdot [\delta]$ in $\pi_1({\mathbb C}^*, z)$ equals $[\sigma]$. Hence, by Lemma 1, at most one of the
both loops $\gamma,\delta$ can be null-homotopic in ${\mathbb C}^*$. Let $\zeta$ denote a simple arc in $\bar{D}\setminus \{0\}$ connecting $p, q$ and otherwise disjoint from the boundary circle (most of the time you can take $\zeta$ to be the Euclidean line segment $pq$, except when $p=-q$). Then $J=f(\zeta)$ is a Jordan curve in ${\mathbb C}^*$. Hence, it separates ${\mathbb C}$ in two components, one of which, called $V$, contains $0=f(0)$. But also $\gamma \setminus \{z\}, \delta\setminus \{z\}$ belong to distinct components of ${\mathbb C}\setminus J$. Suppose that $\delta\setminus \{z\}\subset V$. Then $\delta$ is clearly null-homotopic in
${\mathbb C}^*$ (since it is already null-homotopic in the closed disk $\bar{V}$). qed
I will use the notation $\beta_{pq}\subset S^1$ for the arc defined by $p, q$ such that $f(\beta_{pq})$ is
null-homotopic in ${\mathbb C}^*$. Let $\mathring{\beta}_{pq}$ denote the open arc $\beta_{pq}\setminus \{p, q\}$.
Consider the collection ${\mathcal B}$ of all maximal arcs $\beta_{pq}$ as above (i.e. arcs not properly contained in any arc of the same type); this collection is a countable set. Define
$$
B=\bigcup_{{\mathcal B}} \mathring{\beta}_{pq},
$$
the union of open arcs corresponding to the maximal arcs as above. Its complement is a closed subset $A$ of $S^1$ containing no isolated points. Moreover, if $s, t\in A$ are distinct points such that $f(s)=f(t)$ then $s=p, t=q$ are the end-points of one of the maximal arcs above.
Lemma 3. Define the equivalence relation $\sim$ on $A$ by $p\sim q$, whenever $p, q$ are end-points of one of the maximal arcs $\beta_{pq}\in {\mathcal B}$. Then the quotient space $A/\sim$ is homeomorphic to $S^1$.
Proof: Use the Cantor function to construct this homeomorphism. qed
Now, the map $f|A$ descends to an injective continuous map $S^1=A/\sim \to K$. Hence, its image is a Jordan curve $J$ in $K$.
Lemma. $J=f(A)$ separates $0$ and $\infty$ in $S^2$.
Proof. Enumerate the arcs $\beta_{pq}\in {\mathcal B}$ as $\beta_k:= \beta_{p_kq_k}$, $k\ge 1$. Define the following sequence of maps
$f_n: S^1\to K$: Take $f_0:= f$. Given $f_{n-1}$ define $f_{n}$ by replacing $f_{n-1}|\beta_n$ by the constant map equal to
$f(p_n)=f(q_n)$. Note that, by the uniform continuity of $f$, if ${\mathcal B}$ is infinite,
$$
\lim_{n\to\infty} \mathrm{diam}(f(\beta_n))=0.
$$
It follows that the sequence of maps $f_n$ converges uniformly to $f_\infty$, where $f_\infty$ is obtained from $f$ by replacing each restriction $f|\beta_n$ by the constant map equal to $f(p_n)=f(q_n)$. Therefore, there exists $N$ such that for all $n\ge N$,
$f_n$ is homotopic to $f_\infty$, as maps $S^1\to {\mathbb C}^*$. If ${\mathcal B}$ is finite, of cardinality $N$, we simply set $f_\infty=f_N$. On the other hand, by applying Lemma 2 $N$ times, we see that
the loop $f_N: S^1\to {\mathbb C}^*$ is not null-homotopic (since it is homotopic to the loop $f: S^1\to {\mathbb C}^*$). Therefore, the same is true for the loop $f_\infty$. Note that the image of $f_\infty$ is $J=f(A)$. Hence, $J$ separates $0$ and $\infty$ in $S^2$. qed
Thus, we proved:
Theorem. If $C\subset {\mathbb C}$ is a compact ,connected, locally connected subset separating $0, \infty$ in $S^2$, then $C$ contains a Jordan curve also separating $0, \infty$ in $S^2$.
