I'm wondering if the following can be true:
Let Y be a second countable space and
$\pi\_2:Y \times \mathbb{R}\rightarrow\mathbb{R}$ ($\mathbb{R}$ with its usual topology and
$\pi\_2$ the projection onto the second factor)
be a closed map: do these assumptions imply
that Y is compact? (There is no assumption $T\_0$, $T\_1$ or $T\_2$ on $Y$.)


thank you in advance.



Question about closed projection
Let $(y\_n)$ be a sequence in $Y$. Let $A$ be the subset of $Y \times \mathbf{R}$ of all points $(y\_n, \frac{1}{n})$ for $n \in \mathbf{N}$, and let $B$ be its closure.
Then $\pi\_2[B]$ is closed in $\mathbf{R}$, and contains all points $\frac{1}{n}$, so it contains $0$. So for some $y \in Y$, $(y,0) \in B$. Using the countable base we can extract a subsequence of the $(y\_n)$ that converges to $Y$ (as $Y$ is first countable in particular).
We do then need that $y$ is in the closure of all subsequences of $(y\_n)$ as well, which follows in a similar way, otherwise we cannot get (without separation axioms) a convergent subsequence from first countability alone. But this works.


So $Y$ is sequentially compact, which implies that $Y$ is countably compact (in the covering sense; no separation axioms needed) and as $Y$ is also Lindelöf, being second countable, $Y$ is compact. 


