Metrizable topological space $X$ with every admissible metric complete then $X$ is compact How to prove: 

If $X$ is a metrizable topological space and every admissible metric on $X$ is complete then $X$ is compact. 

I was trying with an idea of contradiction and thereby to construct an incomplete metric on $X$. But not able to construct.
 A: Suppose that $\langle X,\tau\rangle$ is metrizable, every compatible metric on $X$ is complete, and not compact; then there is a countably infinite closed, discrete set $D=\{x_n:n\in\Bbb N\}\subseteq X$. Let $d$ be a compatible metric; $d$ is not totally bounded, so we may assume that there is an $\epsilon>0$ such that $d(x_m,x_n)\ge\epsilon$ whenever $m,n\in\Bbb N$ and $m\ne n$, and by scaling the metric $d$ we may assume that $\epsilon=1$. Let $p$ be a new point not in $X$, and let $Y=X\cup\{p\}$. For $n\in\Bbb N$ let
$$U_n(p)=\{p\}\cup\bigcup_{k\ge n}B_d\left(x_k,2^{-(n+1)}\right)\;.$$
Let $\tau'$ be the topology on $Y$ generated by the base $\tau\cup\{U_n(p):n\in\Bbb N\}$; clearly $X$ is an open subspace of $Y$, and $\operatorname{cl}_YU_{n+1}(p)\subseteq U_n(p)$ for each $n\in\Bbb N$, so $Y$ is regular.
To show that $Y$ is also $T_1$, it suffices to show that $\bigcap_{n\in\Bbb N}U_n(p)=\{p\}$. Clearly $x_n\notin U_{n+1}(p)$ for each $n\in\Bbb N$, so suppose that $x\in X\setminus D$. Then $d(x,D)>0$, so we may choose $n\in\Bbb N$ such that $2^{-n}<d(x,D)$, and clearly $x\notin U_n(p)$. Thus, $Y$ is $T_3$. 
$X$ is metrizable, so it has a base $\mathscr{B}=\bigcup_{n\in\Bbb N}\mathscr{B}_n$ such that each $\mathscr{B}_n$ is locally finite. For $n\in\Bbb N$ let $\mathscr{B}_n'=\mathscr{B}_n\cup\{B_n(p)\}$, and let $\mathscr{B}'=\bigcup_{n\in\Bbb N}\mathscr{B}_n'$; then $\mathscr{B}'$ is a $\sigma$-locally finite base for $\tau'$. By the Nagata-Smirnov metrisation $\langle Y,\tau'\rangle$ is metrizable; let $\rho$ be a compatible metric. 
Then $\rho'=\rho\upharpoonright(X\times X)$ is a compatible metric on $X$. However, the sequence $\langle x_n:n\in\Bbb N\rangle$ converges to $p$ in $Y$, so it is $\rho$-Cauchy and therefore $\rho'$-Cauchy, yet it does not converge in $X$. This contradiction establishes the result.
