Show that $d(E,F) > 0$ - Is $E \times F$ is a compact set here? 
Let $d(E,F)=\inf\{|z-w| : z \in E, w \in F\}$ where $E \subset
 \mathbb{C}$ is compact and $F \subset \mathbb{C}$ is closed such that
  $E \cap F = \emptyset$. Show that $d(E,F) > 0$; namely, $\mathbb{C}$ is a normed
  vector space.

It remains only to show that the minimum is in $\{|z-w| : z \in E, w \in F\}$. I think I can use the continuous function $|\cdot| : E \times F \to \mathbb{R}$ to show that $Im(f) = \{|z-w| : z \in E, w \in F\}$ is also compact.
Question : Is $E \times F$ a compact set here? Otherwise, how can I show that the minimum is in $\{|z-w| : z \in E, w \in F\}$?
 A: In a metric space $(X,d)$, the function $d:X\times X\to \Bbb R$ is notoriously continuous. Given a fixed non-empty subset $F\subseteq X$, the function \begin{align}&d(\bullet, F):X\to \Bbb R\\ &d(x,F)=\inf\,\{d(x,y)\,:\, y\in F\}\end{align} is continuous as well. By definition $d(E,F)=\inf\,\{x\in E\, :\, d(x,F)\}$, which, in your case, is actually a $\min$ because $E$ is compact.
Hence, there exists $\overline x\in E$ such that $d(\overline x,F)=d(E,F)$.
Hence, by definition of $d(\bullet, F)$, there is a sequence $\{y_n\}_{n\in\Bbb N}\subseteq F$ such that $d(\overline x, y_n)\searrow d(E,F)$.
Now, we use the fact that $X=\Bbb C$ and $d(x,y)=\lvert x-y\rvert$. The sequence $\{y_n\}_{n\in\Bbb N}$ is contained in the closed ball $H=E(\overline x,d(\overline x,y_1))$, centered in $\overline x$ with radius $d(\overline x,y_1)$. $H$ is closed and bounded, therefore sequentially compact. Therefore, $\{y_n\}_{n\in\Bbb N}$ has a convergent subsequence $y_{n_k}\to \overline y$.
It follows that $d(\overline x,\overline y)=\lim\limits_{k\to\infty}d\left(\overline x,y_{n_k}\right)=d(E,F)$. Since $F$ is closed, $\overline y\in F$.
