As a bonus, is every compact, $T_0$, second-countable space the image of $2^\omega \times \omega$?
As a further bonus, can we strengthen "image" to "quotient"?
My motivation for these questions comes from an interest in "explicitly-represented" spaces (my notion of what that should mean has developed since asking the linked question, but not enough to give a single formal definition) - hence the tags. If desired, I can elaborate on how I feel each tag is relevant.
We can built a counterexample by adding a bottom element to $\mathbb{N}^\mathbb{N}$. Let $\mathbb{N}^\mathbb{N}_\bot$ have the underlying set $\mathbb{N}^\mathbb{N} \cup \{\bot\}$, and let a set be open if is either an open subset of $\mathbb{N}^\mathbb{N}$ or the entire set.
The space $\mathbb{N}^\mathbb{N}_\bot$ is Quasi-Polish, so in particular sober and countably-based. It is (trivially) compact, because every open cover needs to contain $\mathbb{N}^\mathbb{N} \cup \{\bot\}$.
However, if $s : 2^\mathbb{N} \times \mathbb{N} \to \mathbb{N}^\mathbb{N}_\bot$ were a continuous surjection, then $s^{-1}(\{\bot\})$ is closed in $2^\mathbb{N} \times \mathbb{N}$, so restricting $s$ would give us a continuous surjection from an open subset of $2^\mathbb{N} \times \mathbb{N}$ to $\mathbb{N}^\mathbb{N}$. But every open subset of $2^\mathbb{N} \times \mathbb{N}$ is a countable union of compact sets, thus the same would hold for its image under $s$. But $\mathbb{N}^\mathbb{N}$ is not a countable union of compact sets.
