Pseudocompactness is continuous image-closed

This article gives the statement, and possibly proof, of a topological space property (i.e., pseudocompact space) satisfying a topological space metaproperty (i.e., continuous image-closed property of topological spaces)
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Statement

Suppose $X$ is a topological space and $f:X \to Y$ is a continuous map. Then, if $X$ is a pseudocompact space, the image $f(X)$ (endowed with the subspace topology from $Y$) is also a pseudocompact space.

Proof

Given: A topological space $X$ such that, for any continuous map $g:X \to \R$, $g(X)$ is bounded. A continuous map $f:X \to Y$.

To prove: For any continuous map $h:f(X) \to \R$, $h(f(X))$ is bounded.

Proof: Note first that since $f:X \to Y$ is continuous, so is $f:X \to f(X)$. Henceforth, we think of $f$ as a map from $X$ to $f(X)$.

Consider a continuous map $h:f(X) \to \R$. Define $g = h \circ f$. Then, $g:X \to \R$ is a map. Since both $h$ and $f$ are continuous, so is $g$, and by assumption, we therefore have $g(X)$ bounded. But $g(X) = h(f(X))$ by definition, so $h(f(X))$ is bounded, completing the proof.