Why does Totally bounded need Complete in order to imply Compact? Why does "Totally bounded" need "Complete" in order to imply "Compact"?
Shouldn't the definition of totally bounded imply the existence of a convergent subsequence of every sequence?
No, total boundedness of $\langle X,d\rangle$ implies that every sequence in $X$ has a Cauchy subsequence. Completeness of $\langle X,d\rangle$ implies that every Cauchy sequence in $X$ actually has a limit point in $X$ and therefore converges. The two together therefore imply that every sequence in $X$ has a convergent subsequence, i.e., that $X$ is sequentially compact. Finally, there is a theorem that a metric space is sequentially compact if and only if it’s compact, so total boundedness plus completeness imply compactness.
The confusion is understandable. We restate the two definitions:

Total boundedness says that every uniform open cover has a finite subcover.
Covering compactness says that every open cover, including non-uniform covers, have a finite subcover.

(In metric spaces, open balls can be used in place of open sets/covers.)
A more illustrative counter-example would be $(0, 1)$. It is not compact because I can find an open cover that has no finite subcover, namely
$$B(1, {1 \over 2}) \cup B({1 \over 2}, {1 \over 3}) \cup \cdots = \bigcup_{n = 1}^{+ \infty} B({1 \over n}, {1 \over n + 1})$$
If we stop the construction of the union at any finite $n$, the interval $(0, {1 \over n} - {1 \over n + 1}]$ won't be covered.
It is totally bounded, though. There exists a finite, positive real number, namely $1$, such that, $\forall x, y \in (0, 1)$, $\ d(x, y) &lt; 1$, making $(0, 1)$ bounded. On the real line, bounded is equivalent to totally bounded.