Proof of the description of the closure in a metric space

Proof

We prove each direction separately.

1. Let $(X, d)$ be a metric space, let $S \subseteq X$ and let $x \in X$. Let $\mathcal{F}$ be the family of all closed sets $F$ of $X$ with $S \subseteq F$.
2. By the definition of the closure, $\overline{S}$ is the intersection of the members of $\mathcal{F}$.
3. For $\varepsilon > 0$, let $B(x, \varepsilon)$ be the open ball. A point $s$ lies in $B(x, \varepsilon)$ exactly when $d(x, s) < \varepsilon$.
4. First let $x \in \overline{S}$. Suppose, for a contradiction, that some $\varepsilon > 0$ has no $s \in S$ with $d(x, s) < \varepsilon$. Then $S \cap B(x, \varepsilon) = \varnothing$.
5. Put $F = X \setminus B(x, \varepsilon)$. Its complement $B(x, \varepsilon)$ is open, because open balls are open. So $F$ is closed.
6. By step 4, $S \subseteq F$. Hence $F \in \mathcal{F}$, and $\overline{S} \subseteq F$ by step 2.
7. But $d(x, x) = 0 < \varepsilon$, so $x \in B(x, \varepsilon)$ and $x \notin F$. This contradicts $x \in \overline{S} \subseteq F$.
8. So for each $\varepsilon > 0$ there is $s \in S$ with $d(x, s) < \varepsilon$.
9. Now suppose that for each $\varepsilon > 0$ there is $s \in S$ with $d(x, s) < \varepsilon$. Let $F \in \mathcal{F}$.
10. Suppose, for a contradiction, that $x \notin F$. Then $x$ lies in $X \setminus F$, which is open because $F$ is closed.
11. By the description of the open sets of a metric space, there is $\varepsilon > 0$ with $B(x, \varepsilon) \subseteq X \setminus F$.
12. By hypothesis, there is $s \in S$ with $d(x, s) < \varepsilon$. Then $s \in B(x, \varepsilon) \subseteq X \setminus F$, so $s \notin F$.
13. This contradicts $s \in S \subseteq F$. Hence $x \in F$.
14. So $x$ lies in each member of $\mathcal{F}$, and $x \in \overline{S}$ by step 2.