$\mathbb{R}$ with the right topology is pseudocompact.

Let $f: X \to Y$ be continuous, where $Y$ is a Hausdorff space. Then $f$ is constant.

Proof: if $f(x_1) \neq f(x_2)$ for some two points $x_1$ and $x_2$, then let $O_1$ and $O_2$ be disjoint open neighbourhoods in $Y$ for $f(x_1)$ resp. $f(x_2)$. Then $(x_1 \in) f^{-1}[O_1]$ is non-empty and open (by continuity) and $f^{-1}[O_2]$ also is non-empty and open and $$f^{-1}[O_1] \cap f^{-1}[O_2] = f^{-1}[O_1 \cap O_2] = f^{-1}[\emptyset]= \emptyset$$

But this is a contradiction as two non-empty open subsets of $X=\Bbb R$ in the right-topology always intersect. (it's anti-Hausdorff, as it were).

As the reals in their usual topology is certainly a Hausdorff space, and all singletons are bounded, this immediately implies that $\Bbb R$ in the right topology is pseudocompact.

First thing to note is that a general open set in $X$ is of one of the following types $$ \emptyset, \quad X, \quad (a, \infty). $$ This is because $$ \bigcup_{i\in I} \ (a_i, \infty) = (\ \inf_{i\in I} a_i, \infty).$$ This means that the closed sets are of the form $$ \emptyset, \quad X, \quad (-\infty, b]. $$ Now if you have a continuous map $f: X \rightarrow \mathbb{R}$, then the preimages of of closed sets are closed. Now we have for $x\in \mathbb{R}$ $$ f^{-1}(\{x\}) = (-\infty, b_x] \quad \text{or} \quad f^{-1}(\{x\})= \emptyset \quad \text{or} \quad f^{-1}(\{x\})=X.$$ If $f^{-1}(\{x\})=X$, then $f$ is constant and thus bounded. We just need to rule out that the first case can happen. Assume that there exists $x\in \mathbb{R}$ such that $$ f^{-1}(\{x\}) = (-\infty, b_x]. $$ As $$ \bigcup_{x\in \mathbb{R}} f^{-1}(\{x\}) = X $$ there must exist $y\in \mathbb{R}\setminus \{ x \}$ such that $f^{-1}(\{y\}) = (-\infty, b_y]$. This yields a contradiction as we cannot map a point to two different values. Hence, all continuous functions $g: X \rightarrow \mathbb{R}$ are constant and thus bounded.