If $A$ and $B$ are compact sets in $M$, show that $A\cup B$ is compact
Could someone please check my proof for this problem? My working definition of "compact" is
$P$ is compact iff every sequence in $P$ has a subsequence that converges to a point in $P$.
My attempt:
Let $(x_n)$ be a sequence in $A\cup B$.
$x_n\in A\cup B$
$\implies x_n\in A$ or $x_n\in B$
If there are infinitely many terms of the sequence in $A$, these terms would form a subsequence of $(x_n)$. Let $x_{n_i}\in A$ where $n_i$ increases with $i$. Then $(x_{n_i})$ forms a sequence in $A$. Since $A$ is compact, $(x_{n_i})$ has a subsequence that converges to a point in $A$. Since $(x_{n_i})$ is itself a subsequence of $(x_n)$, we may conclude that $(x_n)$ has a subsequence that converges to a point in $A$. So, $(x_n)$ has a subsequence that converges to a point in $A\cup B$.
Suppose there are only finitely many terms of the sequence in $A$. This would imply that there are infinitely many terms of the sequence in $B$. We then mimic the argument in the previous paragraph to conclude that $(x_n)$ has a subsequence that converges to a point in $A\cup B$.
Thus, in either case, $(x_n)$ has a subsequence that converges to a point in $A\cup B$. So, $A\cup B$ is compact.
QED