Problem. If $M$ is a metric spaces, let $A$ and $B$ be compact, disjoints and non-empty subsets of $M$. Prove that there is $a_{0} \in A$, $b_{0} \in B$ such that $$d(a_{0},b_{0}) \leq d(a,b)$$ for all $a \in A$ and $b \in B$.
Proof. The function $d: A \times B \to \mathbb{R}$ given by $(a,b) \mapsto d(a,b)$ is continuous and $A \times B$ is compact. So, $d(A \times B)$ is compact and so, assumes a minimum, that is, there is $(a_{0},b_{0})$ such that $$d(a_{0},b_{0}) \leq d(a,b).$$
Now, I know that is enough $B$ closed. I trying to prove it.
Second proof. Write $X = \{d(a,b) \mid a \in A, b \in B\}$. Since "$X$ is non-negative" there is $x \in \mathbb{R}_{+}$ such that $x = \inf X$. Thus, we can take a sequence $d((a_{n}),(b_{n})) \to x$. Moreover, there is a subsequence $(a_{n_{k}}) \to a_{0} \in A$. Then $(b_{n_{k(\ell)}}) \to x - a_{0}$ (WLOG). We have $d((a_{n_{k(\ell)}}),(b_{n_{k(\ell)}})) \to x$. Then $a_{0}, b_{0} = x-a_{0}$ are enough.
Im insecure about
"Thus, we can take a sequence $d((a_{n}),(b_{n})) \to x$. Moreover, there is a subsequence $(a_{n_{k}}) \to a_{0} \in A$. Then $(b_{n_{k(\ell)}}) \to x - a_{0}$ (WLOG). We have $d((a_{n_{k(\ell)}}),(b_{n_{k(\ell)}})) \to x$. Then $a_{0}, b_{0} = x-a_{0}$ are enough."
Probably made some mistake. Can someone help me? I just want to know if the second proof is incorrect or not. If yes, I want to know if I really can change the hypothesis "$B$ compact" for "$B$ closed".
If you want to prove this result using only that $A$ is compact and $B$ is closed (in $M$), then you'll need some additional information about the metric space $M$. Otherwise, there are counterexamples like the following. Let $M$ be the subspace $\{0\}\cup(1,2]$ of $\mathbb R$ (with the usual metric). Take $A=\{0\}$, which is compact, and $B=(1,2]$, which is closed in $M$ (though not in $\mathbb R$).