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$\newcommand{\cl}{\operatorname{cl}}$In Rudin's Principles of Mathematical Analysis, Theorem 2.37 says the following:

If $E$ is an infinite subset of a compact set $K$, then $E$ has a limit point in $K$.

I think this statement can be proved even easier than given in the book, if we rely on some Theorems proven earlier in the book:

1) Theorem 2.34 says that compact sets of metric spaces are closed. Therefore $K$ is closed.

2) The closure of $E$, $\cl(E)$ is by definition is closed, and since $E$ is infinite, the set of limit points of $E$, $E'$ is nonempty and again by definiton $E'\subseteq \cl(E)$

3) Theorem 2.27 c) says that $\cl(E)\subseteq F$ for every closed set $F$ such that $E\subseteq F$

So $E\subseteq \cl(E) \subseteq K$, and we are done. Are there any flaws in this logic, or this is indeed an alternative proof?

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It is circular. You are using the fact that $E$ has limit points, which is what you are trying to prove.

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I agree with the other answer. You're using the fact that $K$ is sequentially compact (which is what you are trying to show) when you claim that $E$ has limit points. For example, if $K=\mathbb{R}$ (as a metric space which is not compact), then you can take $E=\mathbb{Z}$ which is closed but has no limit points.

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    $\begingroup$ Thank you. My source of confusion was that, I supposed that infinite subsets of closed sets must have limit points. But this is not the case. $\endgroup$ Mar 10, 2020 at 22:50
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The OP writes

2) The closure of $E$, $cl(E)$ is by definition is closed, and since E is infinite, the set of limit points of $E$, $E'$ is nonempty and again by definiton $E'\subseteq cl(E)$

as a self-contained statement concerning an infinite set $E$ in a metric space.

Let the metric space be $\Bbb R$ and $E = \Bbb Z$.

We have $cl(E) = E$ and $E$ has no limit points. So the OP's 2) argument falls apart.

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