Compact set and metrizable set. A know that the completeness of a metric space $(X,d)$ depend on the metric $d$ i.e. we can have $\mathcal T_d=\mathcal T_{d'}$ but $(X,d)$ complete whereas $(X,d')$ not complete.
Q1) Could someone give me an example of such a fact ? 
Now, I know that compactness doesn't depend on the metric. But I know that in a metric spaces $(X,d)$, a set is compact $\iff$ it's totally bounded and complete. 
Q2) Can I conclude that if $(X,\mathcal T)$ is a compact metrizable space, then $(X,d)$ will be complete for all metric $$d:X\times X\to \mathbb R \ \ ?$$
 A: Q1) $X=\Bbb N$ with $d_1(x,y)=|x-y|$ and $d_2(x,y)=|\frac1x-\frac1y|$.
Q2) Yes, for all metrics which induce the topology $\mathcal T$.
A: Example 1. Let $X =\{1/n:n\in \Bbb N\}.$ For $x,y \in X$ let $d(x,y)=1$ if $x\ne y.$ Let $d'(x,y)=|x-y|.$
Example 2. For $y,z\in \Bbb R$ let $d(y,z)=|y-z|.$ For $x\in \Bbb R$ let $f(x)=\arctan x.$ Then $f$ is a homeomorphism from $\Bbb R$ to the real sub-space $(-\pi/2,\pi/2).$
For $y,z\in \Bbb R$ let $d'(y,z)=d(f(y),f(z))=|(\arctan y)-(\arctan z)|.$  The sequence $(n)_{n\in \Bbb R}$ in $\Bbb R$ is $d'$-Cauchy but has no $d'$-limit.
The idea is a homeomorphic embedding $f:X\to Y$ where $Y$ is metrizable and $f(X)$ is not closed in $Y$. 
A metrizable space is compact iff every metric for it is complete. It is easy to show that every metric for a compact metrizable space is complete.(See Footnote). To show that $any$ non-compact metrizable space has an incomplete metric is $not$ easy.
In case you are not familiar with the terminology, a homeomorphic embedding $f:X\to Y$ is a homeomorphism from $X$ to the sub-space $f(X)$ of $Y.$ 
Footnote: If $(X,d)$ is a metric space and $(p_n)_{n\in \Bbb N}$ is a $d$-Cauchy sequence with no limit point then $\{X\setminus \{p_n:n\geq m\}:m\in \Bbb N\}$ is an open cover of $X$ with no finite subcover.
