Proof of Closure and Interior in a Metric Space: Containments, the Closure is Closed, the Interior is Open

No choice is used. Throughout, membership in an open ball is read off from Open Ball in a Metric Space (ii), and every ball $B_{d}(x,r)$ with $r>0$ contains its centre, since $d(x,x)=0$ (Metric Space (iii)); a subset $F$ is closed in $(X,\mathcal{T}_{d})$ exactly when $X\setminus F$ is open in $(X,d)$, by Closed Subset of a Topological Space (i) and the definition of $\mathcal{T}_{d}$ in the setting. One fact is used repeatedly: open balls are open (a background fact of the setting), so by Open Subset of a Metric Space (i) every point $y\in B_{d}(x,r)$ has a radius $s>0$ with

$$B_{d}(y,s)\subseteq B_{d}(x,r).$$

Proof of (i). If $x\in\operatorname{int}A$, some ball $B_{d}(x,r)$ lies in $A$ (Interior of a Subset of a Metric Space (ii)), and this ball contains its centre, so $x\in A$. If $x\in A$, then every ball $B_{d}(x,r)$ with $r>0$ meets $A$, at $x$ itself, so $x\in\operatorname{cl}A$ (Closure of a Subset of a Metric Space (ii)).

Proof of (iii). Let $x\in\operatorname{int}A$ and choose $r>0$ with $B_{d}(x,r)\subseteq A$. Every $y\in B_{d}(x,r)$ has a radius $s>0$ with

$$B_{d}(y,s)\subseteq B_{d}(x,r)\subseteq A,$$

so $y$ is itself an interior point of $A$. Thus $B_{d}(x,r)\subseteq\operatorname{int}A$, and $\operatorname{int}A$ is open.

Proof of (ii). We show that $X\setminus\operatorname{cl}A$ is open. Let $x\notin\operatorname{cl}A$, so that some ball $B_{d}(x,r)$ does not meet $A$. Every $y\in B_{d}(x,r)$ has a radius $s>0$ with $B_{d}(y,s)\subseteq B_{d}(x,r)$; then $B_{d}(y,s)$ does not meet $A$ either, and $y\notin\operatorname{cl}A$. Hence $B_{d}(x,r)\subseteq X\setminus\operatorname{cl}A$, so the complement of $\operatorname{cl}A$ is open and $\operatorname{cl}A$ is closed.

Proof of (iv). If $\operatorname{cl}A=A$, then $A$ is closed by (ii). Conversely, let $A$ be closed and $x\notin A$. Since $X\setminus A$ is open, some ball $B_{d}(x,r)$ lies in $X\setminus A$; it does not meet $A$, so $x\notin\operatorname{cl}A$. Thus $\operatorname{cl}A\subseteq A$, and equality follows from (i).

Proof of (v). If $\operatorname{int}A=A$, then $A$ is open by (iii). Conversely, if $A$ is open, every $x\in A$ is the centre of a ball $B_{d}(x,r)\subseteq A$, so $A\subseteq\operatorname{int}A$, and equality follows from (i).