Let $K_1,K_2,\ldots,K_N$ be compact subsets of the metric space $(X,d)$. Now I need to show that:
$K_1\cup K_2 \cup \cdots \cup K_N$ is compact.
My Attempt:
I have the definiton: Let $(X,d)$ be a metric space. A subset $K\subseteq X$ is compact when any sequence in $K$ has a convergent subsequence with limit in $K$.
Let ${a_n}$ be a sequence in $K$. $\Rightarrow a\in K_i$ Now i can take a subsequence: $a_{n_j}$ with limit in $a_j \in K_j$ Now i would be able to do this process over and over again.
When aiming for the given definition of (sequence) compactness: Let $(x_i)_{i\in\Bbb N}$ be a sequence in $K_1\cup \ldots \cup K_N$. For each $i\in \Bbb N$ pick $j(i)\in\{1,\ldots,n\}$ such that $x_i\in K_{j(i)}$. Then there must exist $k\in\{1,\ldots,n\}$ such that $j(i)=k$ for infinitely many $i$, i.e., there exists subsequence of $(x_i)_{i\in\Bbb N}$ having all terms in $K_k$. By sequence compactness of $K_k$, there exists a subsequence of this subsequence (so still a subsequence of the original sequence) that converges with a limit in $K_k$ (and hence also in $K_1\cup\ldots\cup K_N$).