It is easy to show that not every pseudompact space is countably compact, i. e. particular point topology on $\mathbb{R}$. $$ \mathcal{P}=\langle \mathbb{R},\tau=\{A\subset\mathbb{R}\mid 0\in A\}\rangle $$
I am wondering if there exists a Hausdorff space with the same properties. I have tried proving that there does not, but got no result, so I think there exists a counterexample.
Definition 1. A topological space $X$ is said to be pseudocompact iff every continuous real-valued function is bounded. By continuous real-valued I mean a countinuous function between topological spaces $X$ and $\mathbb{R}$ with standart topology. It can be shown that every continuous function form $\mathcal{P}$ to $\mathbb{R}$ is constant, hence bounded.
Definition 2. A topological space $X$ is said to be countably compact iff you can choose a finite subcover from any countable open cover of $X$. It is obvious, that $\mathcal{P}$ does not satisfy this definition.