From Topology Without tears:
Prove that each discrete space and each of the spaces $[a,b],(a,b),(a,b],[a,b),(-\infty,a),(a,\infty)\ and\ \{a\}\ for\ a,b\in \mathbb{R}$ with its induced topology is polish space.
Polish spaces:A topological space is said to be Polish space if it is separable and completely metrizable.
I know $\mathbb{Q}\cap [a,b] \subseteq [a,b]$ and is dense in $[a,b]$ and countable so $[a,b]$ is separable and the topology induced by $\mathbb{R}$ is same the topology induced by the euclidean metric on $[a,b]$.So it is metrizable.And since closed subset of complete metric space is complete implies $[a,b]$ is also complete.So,our $[a,b]$ with euclidean metric is separable and completely metrizable hence polish space.
But I am having doubt proving for other spaces: 1)$(a,b)$.This space may be metrizable,separable.But complete?? Suppose Take example of $(0,1)$.I have a sequence $\{\frac{1}{n}\}\subseteq (0,1)$This is Cauchy and converges to zero.But since the $0\notin (0,1)$ the sequence is not convergent in $(0,1)$.So doesn't it imply $(0,1)$ is not complete metric space and hence $(a,b)$.So how $(a,b)$ is polish space??
$(a,b)$ can be given another metric that is complete and induces the same topology.
One such metric can come from a homeomorphism $f: (a,b) \to \Bbb R$, where we introduce the new metric $d(a,a')= |f(a) - f(a')|$ and makes $f$ an isometry from $((a,b), d)$ to $(\Bbb R, d_e)$.
In general, $X$ is Polish iff it can be embedded as a $G_\delta$ (countable intresection of open sets) of a complete separable metric space. So open sets of $\Bbb R$ and the irrationals $\Bbb P$ are also examples of such spaces.