Let $X$ be the subset of the long line consist of rational points with the topology inherits from the long line.


Is $X$ a metrizable space?



Rational points in the Alexandroff line
As stated in the comments, one can show that $\omega\_{1}$ is not paracompact hence non-metrizable. However, one can also give a direct proof that $\omega\_{1}$ is not metrizable without resorting to paracompactness. Here is one way to prooe that $\omega\_{1}$ is not metrizable. Suppose to the contrary that $d$ is a metric inducing the topology on $\omega\_{1}$. The for each limit ordinal $\alpha$ and $n>0$, there is some $f\_{n}(\alpha)<\alpha$ such that
$d(\beta,\alpha)<\frac{1}{n}$ whenever $f\_{n}(\alpha)\leq\beta\leq\alpha$. If $C$ is the set of all limit ordinals, then $C$ is a club set and $f\_{n}$ is regressive. Therefore there is some stationary set $S\_{n}$ and ordinal $\alpha\_{n}$ such that $f\_{n}(\alpha)<\alpha\_{n}$ for each $\alpha\in S\_{n}$. In particular, since $S\_{n}$ in unbounded, whenever $\alpha,\beta>\alpha\_{n}$, there is some $\gamma>\alpha,\beta$ with $f\_{n}(\gamma)<\alpha\_{n}<\alpha,\beta$. Therefore, $d(\alpha,\gamma)<\frac{1}{n}$ and $d(\beta,\gamma)<\frac{1}{n}$, so $d(\alpha,\beta)<\frac{2}{n}$. In particular, if $\alpha=\sup\{\alpha\_{n}|n\in\omega\}$, then whenever $\beta,\gamma>\alpha$, we would have $d(\beta,\gamma)<\frac{2}{n}$ for all $n$, so $d(\beta,\gamma)=0$. This is a contradiction. In particular, since the rational points on the long line contains $\omega\_{1}$, the set of rational points on the long line is not metrizable either.


Another way to prove $\omega\_{1}$ is not metrizable will be to argue that every continuous map $f:\omega\_{1}\rightarrow\mathbb{R}$ is bounded (this is easy), then show $\omega\_{1}$ is not compact (this is also easy). However, any metric space where every continuous real-valued function is bounded is compact. Therefore $\omega\_{1}$ is not metrizable.


