Proof

Let $X$ be a metric space and let $A $ be its subset.

Let $A$ be closed.
Take any convergent sequence $(x_k)_{k\in\mathbb N}$ with $x_k\in A$ for all $k\in\mathbb N$ with $\lim_{k\to\infty} x_k=x$.
Assume, $x\not\in A$.
- Because $X\setminus A$ is open, $X\setminus A$ is a neighborhood of $x$.
- By the definition of convergence there is a $N\in\mathbb N$ such that $x_k\in X\setminus A$ for all $k\ge N$.
- This contradicts $x_k\in A$ for all $k\in\mathbb N$.

Let $x\in A$ for any convergent sequence $(x_k)_{k\in\mathbb N}$ with $x_k\in A$ for all $k\in\mathbb N$ with $\lim_{k\to\infty} x_k=x$. Further let $x\in A$.
We have to show that $A$ is closed.
Assume, $A$ is open.
- Then $A$ is a neighborhood of $x$.
- Therefore there is an $\epsilon > 0$ such that the open ball $B(x,\epsilon)$ is contained in $A$.
- Since $(x_k)_{k\in\mathbb N}$ was an arbitrary sequence, we can choose it in a way that $x$ is contained in the boundary $\delta A$.
- But then, for all $\epsilon > 0$, the open ball $B(x,\epsilon)$ will have points in common with $A$ and with $X\setminus A$.
- This contradicts $A$ being open.
  ∎

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Forster Otto: "Analysis 2, Differentialrechnung im $\mathbb R^n$, Gewöhnliche Differentialgleichungen", Vieweg Studium, 1984