Assume we are in a metric space and $\rho(K,F) =$ inf{$d(x,y): x \in K, y \in F$} $=0$, $K$ closed and $F$ is compact.
Does this necessarily mean we have a sequence in $F$ converging to an element of $K$? I.e., there exist $ \{y_n\} \subseteq F$ and $x_0 \in K$ such that $\lim_{n \to \infty} d(y_n,x_0) = 0$?
would be glad for any help!
Yes, it does.
From the definition of the distance function, for every natural number $n$ you can find $x_n\in K$ and $y_n\in F$ such that $d(x_n,y_n)<\frac{1}{n}$. Since $F$ is compact, the sequence $y_n$ has a convergent subsequence, say, $y_{n_k}\to y_0$ in $F$. Since $d(x_{n_k},y_{n_k})<\frac{1}{n_k}$, we immediately deduce that $x_{n_k}$ also tends to $y_0$. Since $K$ is closed, this implies that $y_0\in K$.
This argument actually shows that if the distance between a compact set and a closed set in a metric space is zero, then $K\cap F\neq\emptyset$.