Let $(X,\mathscr T)$ be a metrizable space such that every metric that generates $\mathscr T$ is bounded. Prove that $X$ is compact.
My attempt:- We know that $(X,\mathscr T)$ is metrizable. So there is a metric on $x$ such that collection of all open sets with respect to the metric is the $\mathscr T$. Let $\{d_\alpha\}_{\alpha \in \Lambda}$ be the collection of metric that generates $\mathscr T$. We know that $\forall \alpha \in \Lambda, (X,d_\alpha)$ is bounded.
Suppose on contrary $X$ is not compact. So $(X,d_\alpha)$ is not sequantially compact. So, there is a $\{x_n\}$ be a sequence in $X$ such that none of its subsequence converges. How do I make contradiction with our assumption?
For $X$ metrisable with this property, if $X$ were not compact, it would not be sequentially compact, so there would be a sequence $(x_n)$ without a convergent subsequence. So $D=\{x_n: n \in \mathbb{N}\}$ is a closed and discrete subspace of $X$ and so we can extends $f(x_n)=n$ from $D$ to $\mathbb{R}$ to a continuous function (Tietze) $F: X \to \mathbb{R}$ (which is clearly also unbounded, as $f$ is).
It's then easy to verify that if $d$ is any compatible metric for $X$ (which exists) then $d_F(x,y)=d(x,y) + |F(x)-F(y)|$ is another compatible metric on $X$ which is unbounded, contrary to the assumption on $X$ that all compatible metrics are bounded. So the assumption that $X$ is not compact was false...