If $\displaystyle E_1$...,$\displaystyle E_n$ are compact, prove that E= $\displaystyle \cup$ i=1 to n $\displaystyle E_i$ is compact.
Give a straightforward proof.
Suppose that the collection $\displaystyle \left\{O_{\alpha}\right\}$ is an open cover for $\displaystyle \bigcup\limits_{j = 1}^n {E_j } $.
For each $\displaystyle m,~1\le m\le n$ we have $\displaystyle E_m\subseteq\bigcup\limits_{j = 1}^n {E_j } $
So $\displaystyle E_m$ is covered by a finite subcollection of $\displaystyle \left\{O_{\alpha}\right\}$.
Now the finite union of finite collections is a finite collection.
That proves that $\displaystyle \bigcup\limits_{j = 1}^n {E_j } $ is compact.
BTW: This proof works in general topological spaces.