If $X$ is a metrizable space such that every metric generating the topology is bounded, then $X$ is compact.
I found a proof by Mr.@Henno Brandsma
suppose to the contrary that $X$ is not compact
Then there is a closed subset $Y$ of $X$ homeomorphic to the integers (a closed and discrete subset, we get this from any sequence without a convergent subsequence.)
Then define an unbounded function $f$ on $Y$ (map the n-th point in the homeomorphism with $N$ to $n$, and extend by Tietze theorem.) $$F: X \to R $$ Then if $d$ is a compatible metric for X, then so is $$d'(x,y) = d(x,y) + |F(x) - F(y)|$$ (a sequence convergent in $d'$ converges in $d$, because $d \le d' $ and one in $d$ converges in $d'$ by continuity of $F$, essentially. Same convergent sequences means same closed sets, so same open sets).
But $d'$ is unbounded. Contradiction !
Therefore $X$ is compact.
My questions :- 1) Why $X$ is not compact gives that there is a closed subset $Y$ of $X$ homeomorphic to the integers ??
2) The Tietze theorem is for continuous function , but I do not know how the unbounded function $f$ will be continuous in this case, could any one give an example for such function ?? Thanks alot.