Maximizing with two arguments with iteration I would like to show that:
$$
\displaystyle\max_{(x,y) \in X\times Y} f(x,y) = \displaystyle\max_{x} \displaystyle\max_{y} f(x,y)
$$
which I obviously believe to be true, although I may be wrong. $X$ and $Y$ are arbitrary choice sets. $f:X \times Y \rightarrow \mathbb{R}$.
My attempt at a proof is below.
Let:
$$
(x^*,y^*) \in \arg \displaystyle\max_{(x,y) \in X\times Y} f(x,y)
$$
Now let's consider the iterated maximization problem. Let:
$$
y^{*}_{I}(x) \subset \arg \displaystyle\max_{y \in Y} f(x,y)
$$
where I use subscript $I$ to denote "iterated". Suppose, to reach a contradiction, that:
$$
x^* \neq x^*_I, \\
x^*_I \equiv \arg \displaystyle\max_{x \in X} f(x,y^*_I(x))
$$
Since:
\begin{gather}
\{x \in X : x \in \arg \displaystyle\max_{x \in X} f(x,y^*_I(x))\} \subset \{x \in X : x \in \arg \displaystyle\max_{(x,y) \in X\times Y} f(x,y)\} \\
\{y \in Y : y=y^*_I(x); x \in \arg \displaystyle\max_{x \in X} f(x,y^*_I(x))\} \subset \{y \in Y : y \in \arg \displaystyle\max_{(x,y) \in X\times Y} f(x,y)\}
\end{gather}
Then:
$$
f(x^*,y^*) > f(x^*_I,y^*_I(x^*_I))
$$
However (this is the crucial step, I can't seem to argue convincingly about it, although it seems obvious to me):
$$
y^* \in \text{Image}(y^*_I(x))
$$
But if this is the case:
\begin{align}
f(x^*,y^*_I(x^*)) & =\displaystyle\max_{x} \displaystyle\max_{y} f(x,y) \\
&=f(x^*_I,y^*_I(x^*_I)),
\end{align}
giving the desired contradiction. Q.E.D.
Thanks in advance.

For any $y$ we have  $f(x,y) \le \sup_{x'} f(x',y)$ and
$\sup_{x'} f(x',y) \le \sup_{y'} \sup_{x'} f(x',y') $. Hence
$\sup_{x,y} f(x,y) \le \sup_{y'} \sup_{x'} f(x',y') $.
In the other direction we have $f(x',y') \le \sup_{x,y} f(x,y)$ for all $x',y'$. Hence we have
$\sup_{x'} f(x,y) \le \sup_{x,y} f(x,y)$ for all $y$ and so
$\sup_{y'} \sup_{x'} f(x',y') \le \sup_{x,y} f(x,y)$ and so they are equal.
If each $\sup$ is attained then we can replace the $\sup$ by $\max$.
