Show that exists $\delta >0$ such that:$d(x,y)\geq \delta$ This is the problem:
Let $X$ be a metric space with metric $d$ and $K\subset X$ compact and $F\subset X$ closed and $K\cap F=\varnothing$.
Let $x\in F,y\in K$.
Prove that there exists $\delta>0$ such that for every $x\in F$ and $y\in K$, $d(x,y)>
\delta.$
My idea was to use sequences,once I know that in a compact set we have a subsequence converging to an element in $K$ and how $F$ is closed there's a sequence converging to an element in $F$.
But ,then I got stuck...I don't know what to do now.Any hint?Much appreciated!

Assume by contradiction, then for every $n$ you can find $x_n\in K, y_n\in F$ such that $d(x_n,y_n)<\frac1n$.
Since $K$ is compact there is a convergent subsequence $x_{n_k}$, whose limit is $x$. By triangle inequality $$d(x,y_{n_k})\leq d(x,x_{n_k})+d(x_{n_k},y_{n_k})$$
Juggle $\varepsilon$'s a bit and derive that $x\in F$, and the contradiction.
