In exercise 43.7 of Johnsonbaugh/Pfaffenberger – Foundations of Mathematical Analysis, we seek to show that if $X \subset M$ is compact, then if $y \in X'$ then there exists a point $a \in X$ such that $d(a,y) \leq d(x,y)$ for all $x \in X$. Moreover, we are asked to show by example that this fails if $X$ is merely closed. I feel as if my proof does not work, since I cannot come up with a an example of a closed subset for which this fails. I've looked at this question, and I think I understand how that works. However, I still fail to see how it would fail if $X$ were closed but not compact.
My "proof":
Suppose the conclusion fails, i.e. for any $a \in X$ there exists $x \in X$ such that $d(a,y) > d(x,y)$. Let $\{x_n\}$ be such a sequence in $X$. Then, since $X$ is compact, $\{x_n\}$ has a convergent subsequence, $\{x_{n,j}\}$. Since $X$ is compact, $lim_{j \to \infty}x_{n,j} = b$ is in $X$. But then $d(b,y) \leq d(x_{n,j},y)$ for all $x_{n,j}$. So we have the conclusion.
Best Answer
Consider $M=[-1,0)\cup(0,1]$. Let $Y=\{-1\}$ and $X=(0,1]$. Note that $X$ is closed in $M$ since $[0,1]\cap M=X$. Now for $y\in Y$, $d(X,y):=\inf_{x\in X} d(x,y)=1$ while $d(x,y)>1$ for any fixed $x\in X$. Thus, there is no $x\in X$ for which the minimum distance is obtained.
The fatal flaw in your proof to being able to generalize to closed sets is that you are using the fact that in a metric space, a sequence in a compact set always has a convergent subsequence. This is not true if the set is merely closed.