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$K \subset \mathbb{R}^n$ compact. $A_i \subset \mathbb{R}^n$ open.

$K \subset \bigcup_{i=1}^{\infty} A_i$, then can one find $m \in \mathbb{N}$, $K \subset \bigcup_{i=1}^{m} A_i$?

I believe I need to use the fact that every bounded sequence has convergent subsequence. But I am unsure, how to exactly use this.

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    That is the definition of compact set. Maybe if you add a few more words and say where exactly you have a problem or you got stuck, people can help you with your question. – 2017-01-27
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    @JMoravitz: Your example doesn't satisfy the assumptions. No, compactness is not about choosing the open cover yourself. The definition is that every open cover has a finite subcover. – 2017-01-27
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    What exactly is your definition of compactness? That every sequence in $K$ has a convergent subsequence? – 2017-01-27

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If you'd like, you can use the fact that $\mathbb{R}^n$ is a metric space, and thus sequentially compact is equivalent to compact.

Proofs of this result can be found in any basic topology book. (Or for example here: http://people.clas.ufl.edu/mjury/files/sequential_compactness_notes.pdf )

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I'm also unsure how to use this fact directly. It would be much easier to prove this statement using the definition of compact sets via open covers: $K$ is compact if every open cover has a finite subcover. Applied here, it tells us that the given open cover of $K$ by the sets $\{A_i\}_{i=1}^{\infty}$ has a finite subcover $\{A_{i_k}\}_{k=1}^n$, i.e. that $K\subseteq\bigcup\limits_{k=1}^n A_{i_k}$, for some indices $i_1