First of all, you're missing the assumption that $(X,d)$ is a complete metric space. If we don't assume that, the statement is false, as for $X = \mathbb{Q}$ (usual metric) we have that $(0,1) \cap \mathbb{Q}$ is totally bounded and its closure $[0,1] \cap \mathbb{Q}$ is not compact.
Fact: if $E$ is totally bounded then so is $\overline{E}$.
(Proof: suppose that $r>0$ has been given. Then $E$ is covered by finitely many open balls $B(x_i,\frac{r}{2}), i=1,\ldots n$. Then certainly $E \subseteq \cup_{i=1}^n D(x_i, \frac{r}{2})$, where $D(x,s) = \{y \in X: d(y,x) \le s\}$ is a closed ball. The finite union of closed balls is closed, so $\overline{E} \subseteq \cup_{i=1}^n D(x_i, \frac{r}{2}) \subseteq \cup_{i=1}^n B(x_i,r)$. As $r>0$ was arbitary, $\overline{E}$ is totally bounded.)
So we know that $\overline{E}$ is totally bounded and complete (as a closed subset of the complete $(X,d)$). So $\overline{E}$ is compact (every sequence has a Cauchy subsequence, which converges, and so we have sequential compactness).
Relatively compact always implies totally bounded (a sequence in $E$ has a convergent subsequence with limit in $\overline{E}$ by relative compactness, and a convergent subsequence is a Cauchy subsequence, so $E$ is totally bounded), but for the reverse implication we really need completeness of $(X,d)$, as I showed above.
In this post I give an argument on why $X=\{0,1\}^I$ is such a space, where $I=\{0,1\}^{\Bbb N}$. This is a huge power of a two point discrete space, so compact Hausdorff and has a sequence without a convergent subsequence even though every net (so every sequence too) has a cluster point, i.e. a convergent subnet. It's a clean(ish) diagonal argument.
$\beta \Bbb N$ (the Cech-Stone compactification of the integers) is another such compact Hausdorff space, where the sequence $x_n =n$ has uncountably many cluster points, but no convergent subsequence at all. It requires a bit more advanced theory though, but if you happen to know the space already..
Best Answer
Proposition: If $(x_\alpha)_{\alpha \in A}$ is a net in $X$ with cluster point $y$, then $y$ is in $\overline{ \{ x_\alpha | \alpha \in A\}}$.
Proof: Suppose $y \not \in \{ x_\alpha | \alpha \in A\}$. We will then show it is a limit point. Let $U$ an open set containing $y$. Then since $y$ is a cluster point there exists some $\alpha \in A$ s.t. $x_\alpha \in U$. So $\{ x_\alpha | \alpha \in A\} \cap U = \{ x_\alpha | \alpha \in A\} \cap U \backslash\{y\} \neq \emptyset$. So $y$ is a limit point.