Let $(X,d)$ be a metric space and $Y_1,\ldots,Y_n \subseteq X$ compact subsets. Then I want to show that $Y:=\bigcup_i Y_i$ is compact only using the definition of a compact set.
My attempt: Let $(y_n)$ be a sequence in $Y$. If $\exists 1 \leq i \leq n\; \exists N \in \mathbb N \; \forall j \geq N\; y_j \in Y_i$ then $(y_n)$ has a convergent subsequence because $Y_i$ is compact. Otherwise,
$$
\forall 1 \leq i \leq n \; \forall N \in \mathbb N\; \exists j \geq N\; y_j \notin Y_i
$$ Assuming for the moment that $n = 2$ and using induction later we have that
$$
\forall N \in \mathbb N \; \exists j \geq N \; y_j \in Y_1 \backslash Y_2
$$ With this we can make a subsequence $\bigl(y_{n_j}\bigr)_{j=0}^\infty$ in $Y_1 \backslash Y_2$. This sequence lies in $Y_1$ and thus has a convergent subsequence. This convergent subsequence of the subsequence will then also be a convergence subsequence of the original sequence. Now we may use induction on $n$.
Best Answer
Let $\mathcal{O}$ be an open cover of $Y$. Since $\mathcal{O}$ is an open cover of each $Y_i$, there exists a finite subcover $\mathcal{O}_i \subset \mathcal{O}$ that covers each $Y_i$. Then $\bigcup_{i=1}^n \mathcal{O}_i \subset \mathcal{O}$ is a finite subcover. That's it; no need to deal with sequences.