Completeness and total boundedness $\iff$ compactness
$(X,d)$ Totally bounded means that $\forall \epsilon >0 \; \exists n(\epsilon) \in N$ and $\exists x_1\ldots x_n \in X$ such that $X=\cup_{i=1}^{n} B_{\epsilon}(x_i)$
Here I will repeatedly use the result : $(X,d)$ is totally bounded $\iff$ $\forall (x_n)\in X\;\; \exists (x_{n_k}) $ which is Cauchy.
$(\Rightarrow)$ Let $(X,d)$ be complete and totally bounded metric space.
I will show that it is sequentially compact, thereby implying that it is compact.
Let $(x_n)$ be any sequence in $X$. I want to show that it has a convergent subsequence in $X$.
$X$ is totally bounded $\rightarrow \; \exists (x_{n_k}) $ Cauchy subsequence of $(x_n)$
now, $X$ is also complete so that we have $(x_{n_k})\to x_0 $ where $x_0 \in X$. So we have produced a convergent subsequence of $(x_n)$
Hence, it is sequentially compact and thus compact.
$(\Leftarrow)$ Now let $(X,d)$ be compact
Let $(x_n)$ be any sequence in $X$, then it has a convergent subsequence ($X$ is sequentially compact) and hence this subsequence is the required Cauchy subsequence. So $X$ becomes totally bounded.
Let $(x_n)$ be any Cauchy subsequence in $X$. Again by sequential compactness, it has a convergent subsequence (say it converges to $x_0$). So original sequence $(x_n)$ also converges to $x_0$. Hence it is also complete.
Is this correct?