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The square $S := [- R, R] \times [-R, R]$ is a compact subset of $\Bbb R^2$.

An intuitive approach: Let $S$ be not compact then there is an open cover of which there is no finite sub cover of $S$.Now we divide the square $S$ into four smaller squares by joining the pairs of midpoints of opposite sides. One of these square will not have a finite subcover from the given cover. For, otherwise, each of these four squares will have finite subcover so that the union of these subcovers will be a cover for $S$. Thus, $S$ itself will admit a finite subcover.

We then choose one such smaller square and call it $S_1$ . And then repeat the process by subdividing $S_1$ into four squares and choosing one of the smaller squares which does not admit a finite subcover.

In this way we get a sequence of squares $S_n$ such that $S$, dose not admit a finite subcover and the length of sides of $S_n$ gradually decreases.

I think we will get a contradiction from here. Please Help to write a formal proof.

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    $\begingroup$ Consider the top left corners of the squares. These points must converge to a point $p$. From there you can derive a contradiction. $\endgroup$ – Gregory Grant Apr 15 '15 at 17:49
  • $\begingroup$ The argument is basically identical for a closed interval in $\mathbb R^1$. You can find that proof in Armstrong Basic Topology. $\endgroup$ – Gregory Grant Apr 15 '15 at 17:50

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