dist$(x,A)=0$ if and only if $x\in \overline{A}$ (closure of $A$)

Your proof is correct, but it can be simplified: $x$ is in the closure of $A$ iff there exists a sequence of elements of $A$ which converges to $x$ iff $d(x,A)=0$.

Your proof is correct, but observe that this problem is really more of a statement about the infimum of a set of real numbers than about topology:

Recall that the infimum $\alpha$ of a nonempty lower-bounded set $S$ of real numbers is the number such that for every integer $n > 0$ there is an element $s_n\in S$ such that $$ \alpha \le s_n < \alpha + \frac{1}{n}. $$

So for this problem, if $d(x,A) = 0$, then $\inf\{d(x,y):y\in A\} = 0$. Thus for every $n$, there is an element $y_n\in A$ such that $d(x,y_n) < 1/n$. But this is just the definition of a sequence of points $\{y_n\}$ in $A$ that converges to $x$. Hence $x$ belongs to the closure of $A$.

Conversely, if $x$ lies in the closure of $A$, then $x$ is the limit of a sequence of points $y_n\in A$ such that $y_n\to x$. (If $x\in A$, there is nothing to show.) Translating $y_n\to x$ into a statement about real numbers, we glean that for every $\epsilon>0$, and all $n$ sufficiently large, $d(x,y_n) < \epsilon$, so that $d(x,A)=\inf\{d(x,y):y\in A\} = 0$ by the definition of infimum.