This is my proof. I just want to know if it has any problems.
Let $A$ be a nonempty set in the metric space $X$. Define the following functions for a point $x \in X$:
$$D(x,A):=\inf_{a\in A} d(x,a)$$
for some metric $d$. Now, for $n \geq 1$ $$F_n := \{x \in X : D(x,A)<1/n\}$$
If $A$ is compact, and $\mu$ is a Borel measure on $X$, then $\mu(A)=\lim_{x \to \infty} \mu(U_n)$
Proof:
Let $x_n$ be some element in the set $F_n$ for each $n \geq 1$. $$D(x_n,A)<1/n$$ $$D(x_{n+1},A)<1/(n+1)<1/n$$
Therefore, $\forall x_{n+1} \in F_{n+1} \to x_{n+1}\in F_n \to F_{n+1} \subseteq F_n$.
This means that $F_1 \supseteq F_2 \supseteq \ldots$
Now, let $x \in A \to D(x,A)=d(x,x)=0<1/n \ \forall n \to x \in F_n \to A \subseteq F_n \to A \subseteq \cap_{n=1}^\infty F_n$
Conversely, let $x \in \cap_{n=1}^\infty F_n$
Therefore, for all $n$, $D(x,A)<1/n \to D(x,A)\leq 1/n$
If we take the limit on both sides of the inequality as $n\to \infty$ we get $$D(x,A) \leq 0$$
Since $D$ is defined in terms of a metric $d$ then it is also true that
$$0 \leq D(x,A)$$
Thus $D(x,A)=0$ and $x \in A$ so $\cap_{n=1}^\infty F_n \subseteq A$.
Since we have both $\cap_{n=1}^\infty F_n \subseteq A$ and $A \subseteq \cap_{n=1}^\infty F_n$ $$A = \cap_{n=1}^\infty F_n$$
Now, we have proved that $F_1 \supseteq F_2 \supseteq F_3 \supseteq \ldots$ and that $\cap_{n=1}^\infty F_n = A$ and it is trivial that $\mu(F_1) < \infty$
Since $\mu$ is a measure, it follows by continuity from above that $$\mu(A)=\lim_{n\to\infty}\mu(F_n)$$
I was told that if $A$ is not closed, but it is bounded, then there are cases where we have $\mu(A) < \lim_{n\to \infty} \mu(F_n)$ but I can't see how that would be possible.