I try to solve the following exercise, which is Exercise 2.48 in Megginson's An Introduction to Banach Space Theory.
Let $X$ be a set and let $\mathcal{F}$ be a separating family of functions $f : X \to Y_f$, where $Y_f$ is Hausdorff for each $f \in \mathcal{F}$. Prove that $X$ is compact with respect to the initial topology induced by $\mathcal{F}$ if, and only if, the image $f(X)$ is compact in $Y_f$ for each $f \in \mathcal{F}$ and the image of $X$ in $ Y := \Pi_{f \in \mathcal{F}} Y_f$ under the map $x \mapsto (f(x))_{f \in \mathcal{F}}$ is closed.
My solution for the forward implication is:
Suppose $X$ is compact. Then the image $f(X)$ is compact in $Y_f$ as the image of a compact set under a continuous function is compact. Recall now that $Y$ is Hausdorff as any $Y_f$ is Hausdorff. Since $x \mapsto (f(x))_{f \in \mathcal{F}}$ is a homeomorphism of $X$ onto a subspace of Y, it is in particular continuous, and hence the image of $x \mapsto (f(x))_{f \in \mathcal{F}}$ is compact in $Y$. As a compact subset of a Hausdorff space is closed, it follows that the image of $x \mapsto (f(x))_{f \in \mathcal{F}}$ is closed in Y.
However, I am not sure how to prove the inverse implication.
Any help or comments are highly appreciated.
If $f(X) \subseteq Y_f$ is compact for all $f \in \mathcal{F}$, then the image of $X \to Y, x \mapsto (f(x))_f$ lies in the compact subspace $\prod_{f \in \mathcal{F}} f(X)$. If this image is closed, then it is compact. Since $x \mapsto (f(x))_f$ is a homeomorphism onto its image, we can conclude that $X$ is compact.