Here's something that I came up with. The proposition below is what you're asking for but I've also encapsulated the main idea behind these results in the following lemma in case that's more helpful.
Lemma: Let $S \subseteq X$ be path-connected and $x^1 \in \overline{S}$. Suppose there exists a countable decreasing (i.e. $U_{i+1} \subseteq U_i$) neighborhood basis $\left( U_i \right)_{i=1}^{\infty}$ in $X$ at $x^1$ such that for each $i$, whenever $s^i \in S \cap U_i$ then there exists a path in $S \cap U_i$ from $s^i$ to some element of $S \cap U_{i+1}$. Then $S \cup \left\lbrace x^1 \right\rbrace$ is path-connected.
Remark: Note that we are not assuming that for all $i$, there exists a path between any two point of $S \cap U_i$. The sets $S \cap U_i$ need not even be connected so this is weaker than requiring local connectivity of $\overline{S}$ at $x^1$.
Corollary: Let $S \subseteq X$ be path-connected. If the condition of the above lemma is satisfied at each $x^1 \in \overline{S}$ (or slightly more generally, if each path-component of the boundary of $S$ contains some point satisfying this condition) then $\overline{S}$ is path-connected.
Prop: Let $S \subseteq X$ be path-connected. Suppose that each path component of $\overline{S} \setminus S$ contains some $x^1$ for which there exists a countable decreasing neighborhood basis $\left( U_i \right)_{i=1}^{\infty}$ in $X$ at $x^1$ s.t. for each $i$ and each path-component $P_i$ of $S \cap U_i$, there exists a path in $\overline{S} \cap U_i$ whose image intersects both $P_i$ and $S \cap U_{i+1}$. Then $\overline{S}$ is path-connected.
Remark: In this proposition, you can replace "of $\overline{S} \setminus S$" with "of the boundary of $S$ in $\overline{S}$". Also, to prove that $\overline{S}$ is path-connected, it may be easier to find some other path-connected $R \subseteq X$ such that $\overline{R} = \overline{S}$ and then apply these results to $R$ in place of $S$.
Proof of lemma: Pick any $s^1 \in S \cap U_1$ and any $0 = t_0 < t_1 < \cdots < 1$ s.t. $t_i \to 1$ and let $\gamma_0 : [t_0, t_1] \to S$ be the constant path at $s^0 := s^1$. Suppose for all $0 \leq l \leq i + 1$ we've picked $s^l \in S \cap U_l$ and for every $0 \leq l \leq i$ we have a path $\gamma_l : [t_l, t_{l+1}] \to S \cap U_l$ from $s^l$ to $s^{l+1}$ (where observe that this holds for $i = 0$). By assumption, we can pick $s^{i+2} \in S \cap U_{i+2}$ and a path $\gamma_{i+1} : [t_{i+1}, t_{i+2}] \to S \cap U_{i+1}$ from $s^{i+1}$ to $s^{i+2}$.
After starting this inductive construction at $i = 0$ we can use $\gamma_0, \gamma_1, \ldots$ to define $\gamma : [0, 1] \to S \cup \left\lbrace x^1 \right\rbrace$ on $[0, 1)$ in the obvious way and then declare that $\gamma(1) := x^1$. For any integer $N$, $l \geq N$ implies $\operatorname{Im} \gamma_l \subseteq U_l \subseteq U_N$ so that $\gamma([t_N, 1]) \subseteq U_N$. Thus $\gamma$ is continuous at $1$ so that $S \cup \left\lbrace x^1 \right\rbrace$ is path-connected. Q.E.D.
It should now be clear how the idea behind this lemma's proof led to the above proposition's statement.
Proof of prop: Let $x^1$ and $\left( U_i \right)_{i=1}^{\infty}$ have the properties described in the proposition's statement, let $0 = t_0 < t_1 < \cdots < 1$ be s.t. $t_i \to 1$, and let $\gamma_0 : [t_0, t_1] \to S \cap U_1$ be any constant path. Suppose $i \geq 0$ is such that for all $1 \leq l \leq i$, we have constructed a path $\gamma_l : \left[ t_l, t_{l+1} \right] \to \overline{S} \cap U_l$ such that $\gamma_l(t_l) = \gamma_{l-1}\left( t_{l} \right)$ and $\gamma_l\left( t_{l+1} \right) \in S \cap U_{l+1}$ (note that this is true for $i = 0$). Our assumption on $\left( U_i \right)_{i=1}^{\infty}$ allows us to construct a path $\gamma_{i+1} : \left[ t_{i+1}, t_{i+2} \right] \to \overline{S} \cap U_{i+1}$ starting that $\gamma_i\left( t_{i+1} \right)$ and ending at some point of $S \cap U_{i+2}$. Exactly as was done in the proof of the above lemma, we may now define a continuous map $\gamma : [0, 1] \to \overline{S}$ such that $\gamma(1) = x^1$. Q.E.D.
Firstly, notice that this doesn't hold for an annulus or donut shape. You take two pieces that go around different sides of the hole and you're done. So somehow the structure of $I^2$ must be relevant (apparently its genus).
So without using more properties of the space, we're going to be stuck. Try drawing some pictures.
Here's one intuitively based approach: let $W$ be $(U\cap V)^c$, the closed subset of $I^2$ not in the intersection. Then $W=U'\cup V'$ where primes indicate restriction to $W$, and $U'\cap V'$ is empty. Hence either one of the primed sets is empty (and we're done as the sets were nested) or $W$ is disconnected.
In this case, use the simply connected property of the original space to note that, picking two random connected components $A,B$ of $U\cap V$, and points $a,b$ within them, any path joining $a,b$ can be continuously deformed into any other. The restriction of these paths to $W$ can also be continuously deformed into each other. But since $W$ is a disconnected union of the primed sets, you can show that each connected component is entirely surrounded by one of the primed sets, contradicting the connectedness of the other.
Edit: Just had a chat with a smarter friend than me, and we concluded that the 'natural' way to do this is homology theory. The logic runs roughly as follows:
$$0 \longrightarrow H_0(U\cap V) \longrightarrow H_0(U) \oplus H_0(V) \longrightarrow H_0(U\cup V) \longrightarrow 0$$
is a short exact sequence; since $U,V,U\cup V$ are connected, all their $H_0$s are $\mathbb Z$. But then since the sequence is exact (and in particular we have an injection for the second arrow) and since the kernel of the addition map $\mathbb Z \oplus \mathbb Z \to \mathbb Z$ is isomorphic to $\mathbb Z$, we conclude that $H_0(U\cap V) \cong \mathbb Z$ and therefore $U\cap V$ is also connected.
Simplifying this argument by stripping out all the general homological algebra would probably result in basically approximating $U,V$ by unions of simplices (with a deeply tedious argument) and then doing some (Euler-characteristic-style) counting to show that the intersection must be connected.
Note: One thing that's worth pointing out is that both the above arguments move to path-connectedness which is in general stronger than connectedness. However open subsets of $\mathbb R^n$ have the property that they are path connected if and only if they're connected. (Proof: Consider path components; show they are both closed and open. This contradicts connectedness.)
Best Answer
It is unnecessary. Your informal proof is correct. See Rotman's Theorem 1.15.
Pick $c \in A \cap B$. Then each $a \in A$ satisfies $a \sim c$ and each $b \in B$ satisfies $b \sim c$, thus each $x \in X$ satisfies $x \sim c$. Since $\sim $ is transitive, any two points $x, y \in X$ satisfy $x \sim y$.