A counter example, topology question

Im looking for a counter example for the following problem:

Let $ S_{1}\neq\emptyset $ be a compact set and $S_{2}\neq\emptyset $ a closed set.(Those are subsets of $ \mathbb{R}^n $ for some $ n $).

So there exists $ s_{1}\in S_{1},\thinspace\thinspace\thinspace s_{2}\in S_{2} $ such that for the distance between the sets:

$ d\left(S_{1},S_{2}\right):=\inf_{x_{1}\in S_{1},x_{2}\in S_{2}}d\left(x_{1},x_{2}\right) $

we have $ \inf_{x_{1}\in S_{1},x_{2}\in S_{2}}d\left(x_{1},x_{2}\right)=d\left(s_{1},s_{2}\right) $

Originally this was a prove/disprove question, but Im pretty sure that this is false, so Im looking for a counter example. Thanks in advance.

After reading the comments:

This is my attempt to prove the statement (rather then disprove it):

Let $ d\left(x_{n}^{1},x_{n}^{2}\right)$ be a sequence such that $ \lim_{n\to\infty}d\left(x_{n}^{1},x_{n}^{2}\right)=\inf_{x_{1}\in S_{1},x_{2}\in S_{2}}d\left(x_{1},x_{2}\right) $.

Since $S_1 $ is compact, there exists a convergent subsequence of $ x_{n}^{1} $, say $ x_{n_{k}}^{1} $.Denote its limit by $ x_1 $. One can verify that

$ d\left(x^{1},x_{n_{k}}^{2}\right)\underset{n\to\infty}{\longrightarrow}\inf_{x_{1}\in S_{1},x_{2}\in S_{2}}d\left(x_{1},x_{2}\right) $

I tried to prove that $ x_{n_{k}}^{2} $ also have a convergent sequence and that would end the proof. But I cant see how.