TITLE: Topology question with closed sets.
QUESTION [2 upvotes]: Let $ K\subseteq \mathbb{R}^n$ be a compact set and let $E\subseteq \mathbb{R}^n$ be a closed set.
***Its also given that $ \inf \{d(x,y)|x\in K, y\in E\}=0$.
$ d(x,y)=\sqrt{\sum_j (x_j-y_j)^2}$

Prove that $E\bigcap K\neq \phi $

So I defined $B$ as the union of $E$ and $K$, and since $E$ and $K$ are closed sets, so is $B$. and from that we obtain that $\mathbb{R}^n \backslash B$ is an open set.
Now I assumed negatively that $ E\bigcap K= \phi $, so I took a point$\{c\}\in\mathbb{R}^n\backslash B$  that is on the shortest vector connecting $2$ points from $K$ and $E$. since $\mathbb{R}^n\backslash B$ is an open set there is a $ r>0 $ s.t the ball $B(\{c\},r)$  is fully contained in $\mathbb{R}^n\backslash B$. 
From that we obtain that the closest distance between $2$ points from $E$ and $K$ is at least $2r (r$ does NOT tend to $0)$, which contradicts ***. Therefore, $E\bigcap K\neq \phi $.
What do you think about this answer?
I got 0 out of 22 points for it. Should I appeal or does this answer make no sense whatsoever?
And if you have another solution it would be nice to see.
Thanks

REPLY [2 votes]: Since the $\inf$ is $0$ you find a sequence $(x_n)_{n\in\mathbb N}$ in $K$ and a sequence $(y_n)_{n\in\mathbb N}$ in $E$ such that
$$\lim_{n\to\infty} d(x_n,y_n)=0.$$
Since $K$ as compact set is bounded the sequence $y_n$ is bounded and hence lies in a compact subset of $E$. This implies that both sequences have a convergent subsequence. The limit of both subsequences have to be the same point because of the continuity of the distance function. Since $K$ and $E$ are closed the limit is element of $K\cap E$, hence the intersection is non-empty.
In your proof you assume to have a shortest vector connecting two points from $K$ and $E$. You can't simply assume that. First you have to show the existence of such a shortest vector. For example if you only know that both sets are closed then the statement is not true. Do you find a counter-example? Your proof doesn't use compactness of $K$ but that is necessary as your counter-example will show us.
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