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!
On
Here is an argument using open covers.
For each $x \in K$ we can find some open ball $B_x$ of positive radius, centered at $x$ and contained in $U$. Let $C_x$ be the open ball centered at $x$ with half the radius of $B_x$. Then $\{C_x : x \in K\}$ is an open cover of $K$, so there is a finite subcover, call it $\mathcal{C}$.
Let $r$ be the smallest radius among the balls in $\mathcal{C}$. I claim that $\epsilon = r$ satisfies the requirements of the problem statement.
To prove this, choose any $p \in K$. Then $p \in C_x$ for some $C_x \in \mathcal{C}$. For any $q$ with $d(p,q) < r$, we have $$d(x,q) \leq d(x,p) + d(p,q) < \text{radius}(C_x) + r \leq 2\text{radius}(C_x) = \text{radius}(B_x)$$ so $q \in B_x$, hence $q \in U$.
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.