Follow up to this question.
I realized that question, which I've asked, explains "why" we can apply Baire's Theorem to $K$. It doesn't address however why $\exists n$ such that $K \cap nE \neq \emptyset$, so this question it's just a check (as I'm reviewing my knowledge of Functional Analysis).
According to Baire's theorem such a $K$ is of second category, which means it is not a countable union of nowhere dense (so it's of second category) or equivalently the countable intersection of open dense in $K$ is not empty.
However I'm not able to reach the conclusion I want (or maybe I'm just not convinced). I guess I can pick a collection of open dense in $K$ therefore (collection is $\left\{ V_i \right\}$)
$$ K = \overline{\bigcap V_i} = \bigcup K \cap nE $$
the bit that is confusing me is when I say "I can pick", can I actually pick such a collection as a consequence of Baire's theorem?
Note that $$K = \bigcup_{n=1}^\infty K \cap nE$$
and for all $n \geq 1$ we have that $K \cap nE$ is closed in $K$ (since $E$ is closed in $X$). By the Baire category theorem (applied to the compact Hausdorff space $K$), there is $n \geq 1$ such that $K \cap nE$ has non-empty interior. In particular, $K \cap nE \neq \emptyset$.