Supremum of Infimum and Infimum of Supremum
Let $f(0,y)=0$ for all $y\ne 0$, and let $f(1,y)=0$ for all $y\ne 1$. Let $f(x,y)=1$ in all other cases. This includes $x=0,y=0$ and $x=1$, $y=1$.
Look at the left-hand side. For any fixed $y$, we have $\inf_{x\in X} f(x,y)=0$. For if $y=0$ we can take $x=1$, and if $y\ne 0$ we can take $x=0$.
Taking the sup of all these $0$'s leaves us at $0$.
Now look at the right-hand side. For any $x$, we have $\sup_{y\in Y} f(x,y)=1$, since for any $x$ there is an $y$ such that $f(x,y)=1$. Taking the inf over all $x$ leaves us at $1$.
Thus the inequality can be strict.
We used the names $0$ and $1$ for certain elements of $X$ and $Y$. They could equally well have been called $x_0,x_1,y_0,y_1$.
Remark: We consider the meaning of an expression such as such as $\sup_{y\in Y}f(x,y)$. The function $f$ is bounded. Fix $x$, and look at $f(x,y)$ as $y$ varies over $Y$. The supremum over all $y$ of $f(x,y)$ is sort of the greatest possible value of $f(x,y)$ for that fixed value of $x$. Not really greatest, it is least upper bound, but for visualization we can think of it as the greatest. So $\sup_{y\in Y} f(x,y)$ is a function of $x$, say $g(x)$. Then, in the expression on the right, we sort of take the smallest possible value of $g(x)$.