Is every boldface $\mathbf{\Pi^0_2}$-definable quotient of Baire space Polish?

Question

I think it's true that every Polish space is a boldface $\mathbf{\Pi^0_2}$-definable quotient of Baire space. However, is the converse true? That is, is every $\mathbf{\Pi^0_2}$-definable quotient of Baire space Polish?
PS: to see that every Polish space is a boldface $\mathbf{\Pi^0_2}$-definable quotient of Baire space, remember that every Polish space is a continuous image of Baire space, and a continuous map is determined by its image on basic open sets, and we can use a real to encode such information as $Distance($image of basic open set $A$, image of basic open set $B)$, and such a real can allow us to define the quotient in a $\mathbf{\Pi^0_2}$ way.
Apologies this is poorly phrased.

Answer 1

No, there are $\Pi^0_2$-definable quotients of Baire space which are not even metrizable.

For example, let $\sim$ be the equivalence relation "$a=b$ or both $a$ and $b$ have infinitely many $1$s". This is $\Pi^0_2$, but in $\mathbb{N}^\mathbb{N}/\sim$ the singleton containing the equivalence class of $(1, 1, 1, 1 , . . .)$ is dense; and no singleton is dense in a metrizable space with more than one point.