Let $X$ be a Hausdorff countably compact space and $Y$ first countably. If $f:X\to Y$ is a continuous bijection then it is a homeomorphism.
Like in the case of compact spaces, I'm trying to show $f$ is closed.
If $A\subseteq X$ is closed, $A$ es countably compact and so is $f(A)$. I want to define a countable open cover for $f(A)$ by using $Y$ is first countably, but I don't know how to do this.
Would you give me a hint?
For this statement to be true, you want to require $Y$ to be Hausdorff, not $X$. Given that correction, the following should be helpful.
Here's one formulation of the proof in the compact case. Suppose you have a point $x$ which is not in $f(A)$. For each $y\in f(A)$, choose disjoint open neighborhoods $U_y$ and $V_y$ containing $x$ and $y$. By compactness, there are finitely many $y_1,\dots,y_n$ such that $V_{y_1},\dots,V_{y_n}$ which cover $f(A)$. The intersection $\bigcap_{i=1}^n U_{y_i}$ is then a neighborhood of $x$ disjoint from $f(A)$. Since $x\not\in f(A)$ was arbitrary, this means $f(A)$ is closed.
To modify this for your situation, try requiring all the sets $U_y$ to be members of some fixed countable base at $x$, and then appropriately combining the sets $V_y$ into only countably many open sets which still cover $f(A)$.