All neighborhoods of a compact subset of an open space are subsets of that open space Let $K$ be a subset of $U$, with $K$ compact and $U$ open. Prove that there is an $\epsilon > 0$ such that for all $p$ in $K$, a neighborhood of radius $\epsilon$ of $p$ is a subset of $U$. Note: This means that the same $\epsilon$ must work for every $P$.
This was a midterm question that I couldn't get and hope to understand before my final, thanks for the help!

Suppose this is false. Then we can choose a sequence $\{x_n\}$ in $K$ such that $B(x_n, 1/n) \not\subset U$. Since $K$ is compact, $\{x_n\}$ has a subsequence $\{x_{n_i}\}$ that converges to some $x \in K$. Since $x \in U$ and $U$ is open, $B(x, 2/m) \subset U$ for some $m$. By making $i$ large enough, we can choose $x_{n_i}$ such that $n_i > m$ and $x_{n_i} \in B(x, 1/m)$. Fix this $i$. We claim that $B(x_{n_i}, 1/n_i) \subset B(x, 2/m)$. Let $y \in B(x_{n_i}, 1/n_i)$, we have
\begin{align}
d(x, y) &\le d(x, x_{n_i}) + d(x_{n_i}, y) \\
&< 1/m + 1/n_i \\
&< 2/m.
\end{align}
Thus, $B(x_{n_i}, 1/n_i) \subset U$. This contradicts the choice of $\{x_n\}$.

Note that this is a special case of Lebesgue's number lemma. The proof of the lemma is essentially the same.
