supremum of expectation $\le$ expectation of supremum?

Edit: sorry - I totally missed you already had the proof in the question - it's correct!

We can equivalently think of this as having a function $f_y$ for each $y$. Then what is always the case is that for each $y$ we have $\sup_y f_y(x)\geq f_y(x)$ for each $x$, and taking the expectation over $X$ this gives $$ \mathbb{E}\left[\sup_y f_y(X)\right]\geq \mathbb{E}\left[f_y(X)\right]$$ Now take the sup over the right side to get the inequality we wanted.

How can you assure $\sup_{y\in \mathcal{Y}} f(X,y)$ is measurable? This is the case if the map $y\mapsto f(x,y)$ is continuous