Metric spaces and limit points question? Let $X, d$ be a metric space. For each $x \in X$ and nonvoid $A, B \in X$, define $$d(x, A) = \inf\{d(x, a) : a \in A\}$$ and $$d(A, B) = \inf\{d(a, b) : a \in A, b \in B\}$$
Prove that $d(x, A) = 0$ implies $x \in \bar A$.
Proof. Suppose $d(x, A) = 0$. Then if we consider $N_\epsilon(x) : \epsilon > d(x, A)$, we get a nonempty intersection $N_\epsilon(x) \cap A$. 
Here's my question: I don't think this is necessarily the case. Suppose $a$ is the only element in $A$. then the intersection is vacuously nonempty. True?
For the converse, suppose $x \in \bar A$. Then $x \in A$ or $N_\epsilon(x) \cap A > \emptyset$. In the former case, $x$ is an interior point. In the latter case, $d(x, a) < \epsilon$. Thus $x$ and $a$ are arbitrarily close and so $x = a$. This completes the proof.
Secondly, any problems with the converse?
 A: In your first "proof" you don't say anything, just conclude what you want to prove. You may to use the infimum characterization:

For every $\varepsilon >0$, there exists $z_{\varepsilon}\in A$ such that
  $$
d(z_{\varepsilon},x)\leq d(x,A)+ \varepsilon
$$

So, in your case you have that for every $\varepsilon >0$, there exists $z_{\varepsilon}\in A$ such that:
$$
d(z_{\varepsilon},x)\leq \varepsilon
$$
Then, taking any ball with radius $\varepsilon >0$, you always have an element $z_{\varepsilon/2} \in A$ in this ball, i.e.:
$$
B(x,\varepsilon)\cap A\neq \emptyset
$$
that is, $x\in \overline{A}$. 
Respect with your proof for the converse, what you say in the case $x\in A$ is false: just think an $x$ in the boundary of a closet set $A$.
And, you always have your second case, namely $\forall \varepsilon >0$ $N_{\varepsilon}(x)\cap A\neq \emptyset$, because of it's the definition of $x \in \overline{A}$ in metric spaces (one of various equivalent definitions)
In order to prove the converse, you can prove with the definition of $x\in \overline{A}$ that:
$$
\forall n\in \mathbb{N}\quad d(x,A)\leq \frac{1}{n}
$$
and conclude taking $n\to \infty$.
