Closure of a region and shortest path 
Let $\Omega$ be a region in $\mathbb{R}^2$ and $\overline{\Omega}$ be closure of $\Omega$. Is it true that between every two points $x,y \in \overline{\Omega}$ exists shortest path (lenth of path is standard path length in $\mathbb{R}^2$).

For example, if $\Omega=\mathbb{R} \setminus (0,0)$ there doesn't exist shortest path between $(0,1)$ and $(0,-1)$ in $\Omega$, but in $\overline{\Omega}$ exists.
 A: This isn't exactly an answer to your question, but one problem is that, even though $\Omega$ is necessarily path-connected (see copper.hat's comment), the same is not necessarily true of $\overline \Omega$. In other words, there may be no path whatsoever between $x,y \in \overline \Omega$, let alone a shortest one.
For an example of this phenomenon, use a "fattened" up version the topologist's sine curve for the open connected set $\Omega$. 

I think that, up to the detail I outlined above, there is always a shortest path. In fact, we have the following.

Proposition 1: Let $X \subset \mathbb{R}^n$ be a closed, path-connected set. Then, for each pair of points $x,y \in X$, there exists a shortest path (in $X$) from $x$ to $y$. 

Proof: Fix $x,y$ in $X$. Let $L$ be the infimum of the lengths of all paths joining $x$ to $y$. Choose a sequence of paths $\{\gamma_n\}$ from  from $x$ to $y$ whose lengths $L_n$ converge to $L$. Moreover, suppose that each $\gamma_n$ is parametrized by arc length so that, in particular, $\gamma_n : [0,L_n] \to X$. Note the domain of each path contains $[0,L]$. Note that, necessarily, $\bigcup \mathrm{range}(\gamma_n)$ is a bounded set (or else $\{L_n\}$ could not be a bounded sequence).
Since, for any fixed point $x \in [0,L]$, the sequence $\{\gamma_n(x)\}$ is bounded sequence, there exists (by the Bolzano-Weierstrass theorem) a subsequence of $\{\gamma_n\}$ such that the corresponding values $\gamma_n(x)$ converge to a point $X$ (recall $X$ is closed). By enumerating the countably many rational points in $[0,L]$ and making a diagonal argument, we find there exists a subsequence $\{\gamma_{n_k}\}$ of $\{\gamma_n\}$ which converges at each rational point $q \in [0,L]$. 
Now, I claim that the sequence of paths $\{\gamma_{n_k}\}$ is uniformly Cauchy on $[0,L]$. First note each of these paths is 1-Lipschitz (with respect to Euclidean distance), as a consequence of the fact that each is parametrized by arc length. Now, for any $\epsilon >0$, find rational numbers $q_1,\ldots,q_n \in [0,L]$ such that every $x \in [0,L]$ is within distance $\epsilon/3$ of some $q_i$. Then find a positive integer $N$ such that $k,\ell > N$ implies that $\| f_{n_k}(q_i) - f_{n_\ell}(q_i)\|<\epsilon/3$ for each $q_i$ in our finite selection. Then, for any point $x \in [0,L]$, there exists an appropriate $q_i$ witnessing
\begin{align*}
 \| f_{n_k}(x) - f_{n_\ell}(x)\| 
&\leq  \| f_{n_k}(x) - f_{n_k}(q_i)\| +  \| f_{n_k}(q_i) - f_{n_\ell}(q_i)\| +  \| f_{n_\ell}(q_i) - f_{n_\ell}(x)\| \\
&\leq \|x-q_i\| + \epsilon/3 + \|x-q_i\| \\
&\leq \epsilon/3 + \epsilon/3+\epsilon/3.
\end{align*}
Since the $\gamma_{n_k}$ are uniformly Lipschitz on $[0,L]$ and the endpoints $L_{n_k}$ of their domains converge to $L$, they converge uniformly to a continuous path $\gamma : [0,L] \to X$ (recalling again that $X$ is closed) from $x$ to $y$. By the lower semicontinuity of arclength (see here) the length of $\gamma$ is less than or equal to $L$, hence has length equal to $L$ by the infimum definition of $L$. 

I'm adding to this answer an improved version of the above proposition. It's an improvement in at least three ways:


*

*The proof has been streamlined by incorporating the suggestions user "copper.hat" made in the comments below.

*The statement has been strengthened slightly in that  a closed path-connected set is no longer assumed. Since the path components of a closed set are not necessarily closed, something is actually gained by this restatement.

*The previous version of the proposition glossed over the fact that, even for a closed path-connected set, there can exist points which aren't joined by a path of finite length. This detail is no longer buried. 



Proposition: Let $X$ be a closed subset of $\mathbb{R}^n$. If $x,y \in X$ are such that there exists some path of finite length from $x$ to $y$, then  there also exists a shortest path from $x$ to $y$. 

Proof: Let $L$ be the infimum of the lengths of all paths joining $x$ to $y$. Choose a sequence of paths $\{\gamma_n\}$ from  from $x$ to $y$ whose lengths $L_n$ converge to $L$. Moreover, suppose that each $\gamma_n$ is parametrized by arc length so that, in particular, $\gamma_n : [0,L_n] \to X$. Note the domain of each path contains $[0,L]$. We shall apply the Arzela-Ascoli Theorem to the sequence of maps $[0,L] \to \mathbb{R}^n$ obtained by restricting the $\gamma_n$ to $[0,L]$. Note the $\gamma_n$ must be uniformly bounded, since $\gamma_n(0)=x$, $\forall n$ and the sequence of lengths $L_n$ is bounded. The $\gamma_n$ are also equicontinuous. Indeed, since each $\gamma_n$ is parametrized by arc length, it is also $1$-Lipchtiz. By the Arzela-Ascoli theorem, there exists a subsequence $\{\gamma_{n_k}\}$ of $\{\gamma_n\}$ which converges uniformly on $[0,L]$ to a path $\gamma : [0,L] \to X$ (recall $X$ is closed) from $x$ to $y$. Since the $\gamma_n$ are $1$-Lipchitz, so is $\gamma$, whence the length of $\gamma$ is less than or equal to $L$. By the infimum definition of $L$, the length of $\gamma$ is then equal to $L$.
