Points of compact metric spaces always have a neighborhood contained in one of the covering sets.

Question

I'm afraid that I might not understand this question very well. I'm probably oversimplifying things, but assuming the contrary holds, i.e. $\forall \ \epsilon >0$, $\exists \ x \in X$ such that $\forall \ i \in I$, $U_\epsilon(x) \nsubseteq U_i$, then wouldn't a $y \in (U_\epsilon(x) \setminus U_i)$ not be covered by the family $U_i$, hence a contradiction?

Answer 1

Your argument is not valid. In order that you can arrive at a contradiction you must have a a $y$ such that $ y \in U_{\epsilon} (x)\setminus U_i$ for every $i$ but there is no such $y$. For a correct proof see https://en.wikipedia.org/wiki/Lebesgue%27s_number_lemma

Answer 2

You correcty negated the statement of the existence of $\varepsilon$ to

$$\forall \varepsilon>0 \exists x \in X: \forall i \in I: \exists p\in X: p \in U_\varepsilon(x) \land p \notin U_i$$

(where $p$ is the non-inclusion "witness")

But as should be clear from this, $p$ depends on $\varepsilon$ and $i$; there is not necessarily a single $p$ that works for all of them. But this is what your supposed $y$ would be....