Theorem. Suppose $(X,d)$ be a metric space.
If $U\subseteq X$ is an open set, then for all elements $a\in U$, there exists a real number $\delta> 0$ such that $B(a;r)\subseteq U$.
I tried to prove the contrapositive using the nested set property:
Proof. Suppose that there exists some element $a\in U$ such that $B(a,\delta)\nsubseteq U$ for all $\delta > 0$. Then it must be true that $B(a,\delta)\cap (X\setminus U)$ is nonempty for all $\delta > 0$. For all $n\in \mathbb{N}$ let
$$A_n=B\Bigl(a,\frac{1}{n+1}\Bigr)\cap (X\setminus U)$$
and for each $n\in\mathbb{N}$ let $x_n\in A_n$, and by the nested set property,
$$\bigcap\limits_{n=0}^{\infty}A_n =\{a\}$$ which shows that $(x_n)_{n\in\mathbb{N}}$ is a sequence in $X\setminus U$ which converges to an element $a\notin X \setminus U$, and therefore $X\setminus U$ is not a closed set.
I'd just like some verification/feedback, or tips on how else this could be proved.