Topology: Proof by Induction. Let $(X,\tau)$ be a topological space and $K_n\subset X$, $n\in \mathbb{N}$ are compact subsets. Show that $\bigcup\limits_{i=1}^{n}K_i$ is compact for every $n\in \mathbb{N}$.
Cutting straight to the inductive step, I have $$\bigcup\limits_{i=1}^{n+1}K_i=\left(\bigcup\limits_{i=1}^{n}K_i\right)\cup K_{n+1}$$ But how I do I know wheter $K_{n+1}$ is compact?
 A: If you have to use induction: by assumption all $K_i$ are compact, so $K_{n+1}$ in particular. The statement we apply induction (in $n$) on in your case is:

($\phi_n$) Let $K_1,\ldots K_n$ be compact subsets of $X$ then $\bigcup_{i=1}^n K_i$ is compact.

So when showing $\phi_{n+1}$ from $\phi_n$, you know that $K_{n+1}$ is compact already (and $\phi_n$ gives you that $\bigcup_{i=1}^n K_i$ is) after which you still have to treat the two-set case as the crucial one.

But you don't need to (re-)use induction at all: suppose $\mathcal{U}$ is an open cover of $\bigcup_{i=1}^n K_i$. For each $1 \le i \le n$, $\mathcal{U}$ also covers $K_i$ and and so there is a finite $\mathcal{U}_i \subseteq \mathcal{U}$ that covers $K_i$ and then $\bigcup_{i=1}^n \mathcal{U}_i$ is a finite subcover of $\mathcal{U}$ for all of $\bigcup_{i=1}^n K_i$. In set theory you show once (by induction) that a finite union of finite sets is finite, and this you can use in all maths areas from then on.
A: Let $K_1$ and $K_2$ be compact subspaces of $X$, and let $\{U_i\}$ be an open cover of $K_1\cup K_2$. Then we have a finite subcover $U_1,\ldots, U_n$ of $K_1$ and a finite subcover $U_{n+1},\ldots, U_m$ of $K_2$. Then $U_1,\ldots, U_m$ is a finite subcover of $K_1\cup K_2$.
