Proof

Let $(X,d)$ be a metric space.
Let the points $x_n$ of a convergent sequence $(x_n)_{n\in\mathbb N}$ be all contained in $X$, i.e. $x_n\in X$, for all $n\in\mathbb N$.
Let the limit $x=\lim_{n\to\infty} x_n$ be also contained in $X$.
We have to show that $U:=\{x_n:~n\in\mathbb N\}\cup \{x\}$ is a compact subset of $X$.
For let an open cover $(U_i)_{i\in I}$ of $U$ be given.
Since $x\in U$ by definition, there is an index $i^*\in I$ with $x\in U_{i^*}$.
Because $U_{i^*}$ is open, $U_{i^*}$ is a neighborhood of $x$.
Because $(x_n)_{n\in\mathbb N}$ converges against $x$, there is an index $N\in\mathbb N$ such that $x_n\in U_{i^*}$ for all $n > N$.
Moreover, every sequence member $x_k$ is covered by some open set $U_{i_k}$ for the finitely many indices $k=0,1,\ldots,N$.
By construction, it follows that \[U\subset U_{i_0}\cup U_{i_1}\cup\ldots\cup U_{i_N}\cup U_{i^*}.\]
Thus, we have found a finite subcover of $U$ in the given open cover $(U_i)_{i\in I}$.
Therefore, $U$ is compact.
∎

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