Let $D := \{ x \in [0, 1]^{\mathbb{N}} \mid \forall n: x_n = 1 \Rightarrow x_{n+1} = 1 \}$ be the set of sequences in $[0, 1]$ such that if $x_n = 1$ for some $n$ then $x_{n+m} = 1$ for all $m \geq 0$.
I know that $[0, 1]^{\mathbb{N}}$ is Polish. Is $D$ with the subspace topology also Polish?
I can represent $D$ as a countable union of pairwise disjoint closed sets $D = (\{ 1 \} \times \{ 1 \} \times \dots) \cup ([0, 1) \times \{ 1 \} \times \{ 1 \} \times \dots) \cup ([0, 1) \times [0, 1) \times \{ 1 \} \times \{ 1 \} \dots) \cup \dots \cup ([0, 1) \times [0, 1) \times \dots)$ but there is no theorem stating that such subsets are necessarily Polish.
I can also represent $D$ as a countable intersection as follows: translate the condition $x_n = 1 \Rightarrow x_{n+1} = 1$ into the set $A = ([0, 1) \times [0, 1]) \cup (\{ 1 \} \times \{ 1 \})$. Then $D = (A \times [0, 1] \times [0, 1] \times \dots) \cap ([0, 1] \times A \times [0,1] \times [0,1] \times \dots) \cap ([0,1] \times [0,1] \times A \times [0,1] \times [0,1] \times \dots) \cap \dots$.
To show that $D$ is Polish it is enough to prove that $A$ is Polish. (Then the sets $A \times [0,1] \times \dots$ and so on are all Polish (as countable products of Polish spaces are Polish) and therefore their countable intersection $D$ is Polish.) $A$ with the Euclidean metric is not complete. Is it possible to construct a compatible complete metric on $A$?
Since $D$ is defined by universal quantifier over a countable set, it is most natural to write it as a countable intersection $\bigcap_n D_n$. Since $G_\delta$ subsets of a Polish space are Polish, it will be enough that $D_n$ is $G_\delta$ for each $n\in\mathbb{N}$.
We have
$$ \begin{align*} D &= \{ x \in [0, 1]^{\mathbb{N}} \mid \forall n: x_n = 1 \Rightarrow x_{n+1} = 1 \}\\ &= \bigcap_n \{ x \in [0, 1]^{\mathbb{N}} \mid x_n = 1 \Rightarrow x_{n+1} = 1 \} \\ &= \bigcap_n \{ x \in [0, 1]^{\mathbb{N}} \mid x_n = 1 \}^{\mathsf{c}} \cup \{ x \in [0, 1]^{\mathbb{N}} \mid x_{n+1} = 1 \}. \end{align*} $$
Now, the first set in the union is open and the second is closed. Their intersection $D_n$ is then $G_\delta$ and we are done.