Proof of Sequential Characterization of the Closure in a Metric Space

Sufficiency. Suppose $(a_m)_{m\in\mathbb{N}}$ is a sequence in $X$ with $a_m\in A$ for every $m\in\mathbb{N}$, converging to $x$ in $(X,d)$. Let $\varepsilon>0$ be a real number. By the definition of convergence there is $N\in\mathbb{N}$ such that $d(a_m,x)<\varepsilon$ for every $m\in\mathbb{N}$ with $m\ge N$; taking $m=N$ gives $d(a_N,x)<\varepsilon$. By condition 3 of the definition of a metric we have $d(x,a_N)=d(a_N,x)$, so $d(x,a_N)<\varepsilon$ with $a_N\in A$. Since $\varepsilon$ was arbitrary, condition 3 of Characterization of the Closure in a Metric Space by Open Balls holds, and that theorem gives $x\in\operatorname{cl}_X(A)$.

Necessity. Suppose $x\in\operatorname{cl}_X(A)$. By Existence of a Sequence of Positive Real Numbers with Limit Zero there is a sequence $(h_m)_{m\in\mathbb{N}}$ of real numbers such that $0<h_m$ for every $m\in\mathbb{N}$ and such that $(h_m)_{m\in\mathbb{N}}$ has limit $0$.

For each $m\in\mathbb{N}$ let $A_m$ be the set of all $a\in A$ with $d(x,a)<h_m$. By condition 3 of Characterization of the Closure in a Metric Space by Open Balls, applied with $\varepsilon=h_m$, the set $A_m$ is nonempty. Since $A\subseteq X$, the assignment sending $m$ to $A_m$ is a family of subsets of $X$ indexed by $\mathbb{N}$, and every member of this family is nonempty. Hence Axiom of Countable Choice provides a sequence $(a_m)_{m\in\mathbb{N}}$ in $X$ with $a_m\in A_m$ for every $m\in\mathbb{N}$. In particular $a_m\in A$ and $d(x,a_m)<h_m$ for every $m\in\mathbb{N}$.

It remains to check that $(a_m)_{m\in\mathbb{N}}$ converges to $x$ in $(X,d)$. Let $\varepsilon>0$ be a real number. Since $(h_m)_{m\in\mathbb{N}}$ has limit $0$, there is $N\in\mathbb{N}$ such that $|h_m-0|<\varepsilon$ for every $m\in\mathbb{N}$ with $m\ge N$. Fix such an $m$. By claim 4 of Additive Cancellation and Elementary Additive Identities in a Field we have $h_m-0=h_m$, so $|h_m|<\varepsilon$, and claim 9 of Properties of the Absolute Value in an Ordered Field, applied in the ordered field of real numbers, gives $h_m<\varepsilon$. By condition 3 of the definition of a metric we have $d(a_m,x)=d(x,a_m)<h_m$, and $h_m<\varepsilon$, so $d(a_m,x)<\varepsilon$ by claim 2 of Elementary Order Arithmetic in an Ordered Field. As $\varepsilon$ was arbitrary, $(a_m)_{m\in\mathbb{N}}$ converges to $x$ in $(X,d)$.