Finite Union of Compact Sets is Compact

Theorem
Let $T = \left({S, \tau}\right)$ be a topological space.

Let $n \in \N$ be a natural number.

Let $\left\langle{U_i}\right\rangle_{1 \mathop \le i \mathop \le n}$ be a finite sequence of compact subsets of $T$.

Let $\mathcal U_n := \displaystyle \bigcup_{i \mathop = 1}^n U_i$ be the union of $\left\langle{U_i}\right\rangle$.

Then $\mathcal U_n$ is compact in $T$.

Proof
Proof by induction:

For all $n \in \N_{> 0}$, let $P \left({n}\right)$ be the proposition:
 * $\mathcal U_n := \displaystyle \bigcup_{i \mathop = 1}^n U_i$ is compact in $T$.

$P \left({0}\right)$ is the case:
 * $\mathcal U_0 := \displaystyle \bigcup_{i \mathop = 1}^0 U_i$

From Union of Empty Set:
 * $\displaystyle \bigcup_{i \mathop = 1}^0 U_i = \varnothing$

From Empty Set is Compact Space it follows that:
 * $\mathcal U_0$ is compact in $T$.

$P \left({1}\right)$ is true, as this just says:
 * $U_1$ is compact in $T$.

Basis for the Induction
$P \left({2}\right)$ is the case:
 * $U_1 \cup U_2$ is compact in $T$.

which is proved as follows:

Let $\mathcal C$ be an open cover of $U_1 \cup U_2$.

Then $\mathcal C$ is an open cover of both $U_1$ and $U_2$.

As $U_1$ and $U_2$ are both compact in $T$:
 * $U_1$ has a finite subcover $C_1$ of $\mathcal C$
 * $U_2$ has a finite subcover $C_2$ of $\mathcal C$.

Their union $C_1 \cup C_2$ is a finite subcover of $\mathcal C$ for $U_1 \cup U_2$.

From Union of Finite Sets is Finite it follows that $C_1 \cup C_2$ is finite.

As $\mathcal C$ is arbitrary, it follows by definition that $U_1$ and $U_2$ is compact in $T$.

This is our basis for the induction.

Induction Hypothesis
Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 2$, then it logically follows that $P \left({k+1}\right)$ is true.

So this is our induction hypothesis:
 * $\mathcal U_k := \displaystyle \bigcup_{i \mathop = 1}^k U_i$ is compact in $T$.

Then we need to show:
 * $\mathcal U_{k+1} := \displaystyle \bigcup_{i \mathop = 1}^{k+1} U_i$ is compact in $T$.

Induction Step
This is our induction step:

We have that:
 * $\displaystyle \bigcup_{i \mathop = 1}^{k+1} U_i = \left({\bigcup_{i \mathop = 1}^k U_i}\right) \cup U_{k+1}$

By the induction hypothesis:
 * $\mathcal U_k$ is compact in $T$.

By the basis for the induction:
 * $\mathcal U_k \cup U_{k+1}$ is compact in $T$.

So $P \left({k}\right) \implies P \left({k+1}\right)$ and the result follows by the Principle of Mathematical Induction.

Therefore:
 * $\displaystyle \forall n \in \N: \bigcup_{i \mathop = 1}^n U_i$ is compact in $T$.