In Milgrom and Roberts (1990) (http://www.uvm.edu/pdodds/files/papers/others/1990/milgrom1990b.pdf), on page 1266, in the proof of Theorem 5, last line (...but then, by (A2)...).
To put their argument in a broader context, they say: Take a function $f:X\times Z \rightarrow \mathbb{R}$, and suppose that $f$ is continuous in $z \in Z$ for fixed $x \in X$, and continuous in $x \in X$ for fixed $z \in Z$, so separately, but not jointly, continuous.
Suppose that $(x_{n},z_{n})_{n \geq 1} \subseteq X \times Z$ is a sequence that converges to some $(x^{\star},z^{\star})$, and that for some $x \in X$,
$$ f(x,z^{\star})>f(x^{\star},z^{\star}). $$
They then make the following argument: Due to the continuity assumptions, there exists some $n$ such that
$$ f(x,z_{n})>f(x_{n+1},z_{n}). $$
Now, for reasons that are clear in the paper, this would be a contradiction. However, disregarding the game theory behind it, I don't see how you make this argument without joint continuity. To see why, let us consider the obvious attempt at applying separate continuity: For $z^{\star}$ fixed, we can use continuity in $x$ to guarantee some $N$ such that for all $k \geq N$,
$$ f(x,z^{\star})>f(x_{k},z^{\star}), $$
this is fine. Now we can apply continuity in the $z$ argument: There exists some $N'$ such that for all $m \geq N'$,
$$ f(x,z_{m})>f(x_{k},z_{m}). $$
However, notice that the $N'$ chosen for the $z$ argument must be done with a fixed $x_{k}$ to begin with. Therefore, if $N'$ is larger than $k$ (if $N'>k$), we cannot guarantee that we can find some $n$ such that
$$ f(x,z_{n})>f(x_{n+1},z_{n}), $$
as required. In fact, it seems to me that if one were able to make such an argument, you would be able to show that separate continuity implies both joint upper and lower semi-continuity, which would imply continuity, which we know is not always true.
Maybe I'm missing something obvious here. If so, any help would be highly appreciated. Thank you!