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I am aware that there is a similar question elsewhere, but I need help with my proof in particular. Can someone please verify my proof or offer suggestions for improvement?

Show that a finite union of compact subspaces of a topological space $X$ is compact.

Let $A_1, \ldots, A_n$ be compact subspaces of a topological space $X$. Let $\mathscr{B}$ be a collection of open sets of $X$ which covers $\displaystyle{\bigcup_{i=1}^n A_i}$. Then, $\mathscr{B}$ covers $A_i$ for each $1 \leq i \leq n$. Since each $A_i$ is compact, we can choose a finite subcover $\mathscr{B}_i$ of $A_i$. But then, $\displaystyle{\bigcup_{i=1}^n \mathscr{B}_i}$ forms a finite subcover of $\displaystyle{\bigcup_{i=1}^n A_i}$.

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    $\begingroup$ That will do... $\endgroup$ – Stefan Mesken Aug 14 '14 at 11:21
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There is nothing wrong with your proof. It´s ok.

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everything is ok ...nice proof

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