Let $(X,\mathscr T)$ be a metrizable space such that every metric that generates $\mathscr T$ is bounded. Prove that $X$ is compact.

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?

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