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I sort of get it, intuitively speaking, but I am hoping someone can write down an answer that clarifies a lot of the subtleties. My go-to example is the interval $(-1, 0) \cup (0, 1)$, and the sequence $(1/(n+1))_{n = 1}^{\infty}$. It is a Cauchy sequence, but not convergent because even when given a very small $\epsilon$, any candidate value for convergence $L$ will get far enough away from $L$ so that $|L - 1/(1 + n)| > \epsilon$.

Okay, but if I consider the usual proof on the reals for the fact that a Cauchy sequence is convergent, I notice that it relies on the Bolzano-Weierstrass (B-W) guaranteeing the existence of a convergent subsequence within a bounded sequence---this gives us a specific value which we can say the Cauchy sequence must converge to.

Every Cauchy sequence is bounded, even in a generalized metric space. Every bounded sequence, in a generalized metric space also contains convergent subsequences, as per B-W. Shouldn't that be enough to latch onto a point of convergence? I am missing some crucial subtlety here...

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The point is that you are wrong: every bounded sequence need not contain convergent subsequences in general.

You can even take a metric which is complete, generates the same topology of a space which has this property, but it isn't true. For instance, take $\mathbb{R}^n$ with the bounded metric $\rho=\min\{1,d\}$. Every sequence is bounded. In particular, $x_n=n$ is bounded, but none of its subsequences converges.

Bolzano-Weierstrass has its generalizations, though. For example, every compact metric space has the property that every bounded sequence has convergent subsequences (however, every sequence is bounded since the space is compact). The generalizations become more palatable if you take away the boundedness assumption. The property that "every sequence has convergent subsequences" is called sequential compactness. For example, every compact first-countable space is sequentially compact, as can be seen here (obs: this example covers the case of compact metric spaces). Every limit point compact first-countable $T_1$ space also has this property, and $T_1$ is necessary, as seen here. Etc

The "boundedness" in the traditional Bolzano-Weierstrass can be seen as just assuming that your sequence is inside a compact set (a $k$-cell, for instance), reducing it to the cases above.

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Aloizio Macedo
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