Was just wondering if my proof was sound. If there is something I should add or make more clear please let me know. Thanks.
Let $A\subseteq \mathbb{R}^{n}$ be closed and $\vec{x}\in \mathbb{R}^{n}$. Define $d(\vec{x},A):= \inf \{\|\vec{a}-\vec{x} \| : \vec{a}\in A\}$.
Show that $\vec{x}\in A \iff d(\vec{x},A) = 0$
($\implies$) Since $A$ is closed, $A$ contains all of it's limit points. Let $\{\vec{a}_{n}\}_{n\geq 1}\subseteq A $ such that $\vec{a}_{n} \to \vec{x}, n \to \infty$
i.e. $\|\vec{a}_{n} - \vec{x} \| \to 0$ as $n\to \infty$
By definition we have $0\leq \inf \{\|\vec{a}-\vec{x} \| : \vec{a}\in A\}\leq \|\vec{a}_{n} - \vec{x} \|$ passing to the limit we have that $d(\vec{x},A):= 0$
$(\impliedby)$ By the property of $\inf$ $\exists \vec{x}_{n}\in A$ such that $\|\vec{x}_{n}- \vec{a}\|\leq d(\vec{x},A)+\frac{1}{n} = \frac{1}{n}$ since $d(\vec{x},A)=0$. Then $\vec{x}_n \in \bar{A} = A$ since $A$ is closed. $\square$
First, you seem to use inconsistent variable naming. In the problem description, you name $\vec{x}$ as the single point whose distance to $A$ you look at. Later, $\vec{x}$ is no longer used but $a$ seems to have taken its place. At the very end, $A$ seems to have been replaced with $V$.
Your ($\implies$) part is needlessly complicated, you don't need to look at any kind of limit or use closeness. You simply have
$$0\leq \inf \{\|\vec{a}-\vec{x} \| : \vec{a}\in A\}\leq \|\vec{x} - \vec{x} \| = 0$$
because $\vec{x}\in A$, and the infimum is at most as big as any value in the set over which it is taken.
Your ($\impliedby$) is only almost done, maybe it is just a typo. You correctly have $\|\vec{x}_{n}- \vec{a}\|\leq \frac{1}{n}$ which implies $\vec{x}_n \rightarrow \vec x$. Because $\vec{x}_n \in A$ and closeness of $A$ you can conclude $\vec x \in A$.