This question has some answers with examples of non-compact pseudocompact Hausdorff (even normal or better, some of them) spaces.
Another nice example (not mentioned in the above answer, which has several) is the following subset of $[0,1]^I$ in the product topology, where $I$ is uncountable: $\Sigma = \{f: I \rightarrow [0,1]: |\{x: f(x) \neq 0 \}| \le \aleph_0 \}$, the so-called Sigma-product of the uncountably many copies of $[0,1]$. This is a proper dense subset of $[0,1]^I$, so not compact, but any real-valued function on $\Sigma$ depends on countably many coordinates, so is essentially defined on a compact set, so is bounded. (details omitted, but can be found online).
As to the realcompactness, this is not very elementary, and expects some background in the Cech-Stone compactification etc. So it will depend on what you already know. But there is one argument, assuming some background. Suppose $X$ is compact. Then $X$ is pseudocompact (because for every real-valued continuous $f$ defined on $X$, $f[X] \subset \mathbb{R}$ is compact, hence bounded) and realcompact (as $X = \beta(X)$ so $X = \nu(X)$ as well, where the latter the real-compactification of $X$). On the other hand, if $X$ is realcompact and $X$ is non-compact, there exists some real-valued continuous function $f$ on $X$ that cannot be extended to $\beta(X)$ (because we then have $X = \nu(X) \subseteq \beta(X), \beta(X)\setminus \nu(X) \neq \emptyset$, and $\nu(X)$ is maximal with the real-valued extension property). But this function then cannot be bounded, as then we could have extended it to $\beta(X)$. So $X$ is not pseudocompact.
