Topology Theorem on closure of a set

Question

If A is a subset of a metric space X ,then A closure is a closed set and is a subset of every closed set containing A

Answer 1

In a metric space, for any subset $A$ we have $A'' \subseteq A'$ (in general spaces this might fail, but it holds in $T_1$ spaces), and a set $B$ is closed iff $B' \subseteq B$ (take this as the definition if you will).

Now defining $\operatorname{cl}{A}=A \cup A'$, as per the comments, we see that $$(\operatorname{cl}{A})' = (A \cup A')' = A' \cup A'' \subseteq A' \cup A' = A' \subseteq \operatorname{cl}{A}$$

so that $\operatorname{cl}{A}$ is closed, and if $C$ is closed and $A \subseteq C$ we have $A' \subseteq C'$ and so

$$\operatorname{cl}{A} = A \cup A' \subseteq C \cup C' \subseteq C$$ as $C' \subseteq C$ because $C$ is closed, proving the final claim.