Question: Let $E$ and $F$ be closed disjoint sets in $\mathbb{R}^d$. Show that if one of the sets is compact, then necessarily $d(E, F) > 0$.
A proof was provided, and I understand it, but I can't figure out why the following 'counterexample' does not work:
Let $E = \left\{\sum_{i=1}^n \frac{1}{2^i}\ |\ n\in \mathbb{N} \right\} $ and $F = \left\{\sum_{i=1}^n \Big(\frac{1}{2^i} + \frac{1}{3^{i+1}}\Big)\ |\ n\in \mathbb{N} \right\} $
Here, both $E$ and $F$ are compact, but $\nexists \epsilon > 0$ such that $d(E,F) > \epsilon$.
Bot $E$ and $F$ are not compact nor closed, since they do not include their limit points $$ \sum_{i=1}^\infty \frac{1}{2^i}\;\text{ and }\;\sum_{i=1}^\infty \left(\frac{1}{2^i}+\frac{1}{3^{i+1}}\right) $$ therefore the theorem does not apply
Note
Following the comments, it should also be noted that $E$ and $F$ are not compact since not any of their open covers contains a finite subcovering: for example, think of the family of open balls $$ \mathcal{C}=\left\{\mathring{B}\big(x_i,10^{-i^2}\big)\right\}_{i\in \mathbb{N}_+} $$ where
$x_i=\frac{1}{2^I}$ or $x_i=\frac{1}{2^i}+\frac{1}{3^{i+1}}$, $i\in \mathbb{N}_+$ and
$\mathring{B}(x_o,r)=\{x\in\mathbb{R}^n|\,|x_o-x|<r\}$.
No one of its finite subfamilies covers the whole $E$ nor $F$.