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Let $X$ be a compact Hausdorff topological space.

Is it true that every sequence has a converging subsequence?

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    Now that from the answers you know about the terminology [sequentially compact space](https://en.wikipedia.org/wiki/Sequentially_compact_space), you can also use [pi-base](http://topology.jdabbs.com/search?q=%7B%22and%22%3A%5B%7B%2216%22%3Atrue%7D%2C%7B%2220%22%3Afalse%7D%5D%7D) to search for examples. – 2017-01-27
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    See also [Compactness / sequentially compact](http://math.stackexchange.com/q/152447) and other posts [linked there](http://math.stackexchange.com/questions/linked/152447). – 2017-01-27

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No, there are compact Hausdorff spaces without any non-trivial (trivial meaning eventually constant) convergent subsequences,like $\beta \omega$, the Cech-Stone compactification of the natural numbers. compact non-sequentially compact spaces like $[0,1]^I$ for $I$ of size $|\mathbb{R}|$. For so-called sequential spaces, which include the first countable and the metric ones, this implication does hold. The Bolzano-Weierstrass peoperty is now called being sequentially compact, BTW.

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"Every sequence has a convergent subsequence" is the property nowadays called "sequentially compact". It is not equivalent to "compact" for Hausdorff spaces. But it is equivalent for metric spaces.