I saw this question posted and came up with a different proof than was answered. I'm hoping someone can review my proof and offer validation/corrections.
Let $X$ be some metrizable space, $U$ some open set in $X$, and $n\in\mathbb{N}$.
Define $$S_n(U):=\left\{x\in X ∣ B\left(x,\frac{1}{n}\right) ⊆ U \right\}.$$ Show that $S_n$ is closed.
Proof:
First, all $x$ must be elements of $U$ or else the open ball about $x$ would not be fully contained in $U$.
So, assume we have some $S_n(U)$ and so there exists some $y\in X - U$. If there exists no such $y$, then $S_n(U)$ is the entire set $X$, and so is thus closed.
We hope to show that there can exist no limit point outside of $S_n(U)$. Hence $S_n(U)$ contains all of its limit points and is then closed.
Suppose $z$ is a limit point of $S_n(U)$ not in $S_n(U)$.
Case one: $z$ is not an element of $U$. Then the open ball of radius $1/n$ around $z$ cannot intersect any $x \in S_n(U)$ since then $z$ would fall in the open ball about $x$ of that same radius, but that radius was given to not contain any points outside of $U$.
Case two: $z$ is an element of $U$. So $B(z,1/n)$ must contain some $y\in X - U$. Then there exists some distance $b = (1/n) - d(z,y)$ that is greater than zero. Then any radius of $b$ about $z$ cannot contain any $x\in S_n(U)$ since if $x$ did reside within distance $b$ from the point $z$ we would have:
$$ d(x,y) \le d(x,z) + d(z,y) $$
But the right-hand side is $< (1/n) - d(z,y) + d(z,y) = (1/n)$.
Hence $y$ would fall within the $1/n$ ball about $x$, a contradiction since $y$ is not a member of $U$.
Hence we can find an open set containing $z$ that does not contain any element of $S_n(U)$. So $z$ is not a limit point. Hence no limit point can belong outside of $S_n(U)$, hence $S_n(U)$ is closed.