Proof
(related to Proposition: Convergent Sequence together with Limit is a Compact Subset of Metric Space)
 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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References
Bibliography
 Forster Otto: "Analysis 2, Differentialrechnung im \(\mathbb R^n\), Gewöhnliche Differentialgleichungen", Vieweg Studium, 1984