A complex-analytic space $X$ is holomorphically convex iff for all compact $K\subset X$, the subset $$\hat{K}:=\{p\in X : |f(p)|\leq \max_{x\in K}|f(x)| \forall f\in\mathcal{O}(X)\}$$ is again compact.
Why should a product of holomorphically convex spaces again be holomorphically convex? I have seen this stated without proof many times, but I am having trouble coming up with a proof. I have made a few attempts at proving this by attempting to say something about the relationship between a subset $K$ and the product of it's projections, but I haven't gotten anywhere with it.
Here's a solution - turns out sleeping on it was enough to figure it out.
$\hat{K}$ is closed for any $K$ in any space, as it's an intersection of closed sets of the form $|f|^{-1}([0,\max_{x\in K} |f(x)|])$. So it suffices to show that $\hat{K}\subset X\times Y$ contained in a compact set, namely $\widehat{pr_X K}\times \widehat{pr_X K}$.
Suppose $(x,y)\in \hat{K}$. Then for all $f\in \mathcal{O}(X\times Y)$, $|f(x,y)|\leq \max_{x\in K} |f(x,y)|$. Specifically, this is true for all functions $f_X$ which are constant in the $Y$ direction, and this gives that $x\in \widehat{pr_X K}$. Similarly, $y\in \widehat{pr_Y K}$, so this shows that $\hat{K}\subset \widehat{pr_X K}\times \widehat{pr_Y K}$, so we are done.