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Let $(x_k)$ be a convergent sequence in $\mathbb{R}$ such that $(x_k) \to x$. Let $A = \{x_1, x_2, \ldots\} \cup \{x\}$.

It's quite easy to show that $A$ is compact by showing that every open cover has a finite subcover. And since $\mathbb{R}$ is a metric space, it follows that $A$ is sequentially compact.

But I need to show directly that $A$ is sequentially compact. So how can I show that every sequence in $A$ has a subsequence that converges in $A$?

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Hint: choose $y_n$ a sequence in $A$. If $y_n$ has only finitely many terms you are done. Otherwise keep taking terms of $y_k$ that correspond to higher and higher indexed terms of $x_n$ to show that $y_k$ has a subsequence converging to $x$.

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