Suppose I have a map $S:X \to Y$ between Banach spaces which is multivalued, so that $S(x)$ is a set.
I have shown that $S$ takes a closed ball of radius $R$ to itself, and it also is such that if $x_n \to x$ weakly, then there exists $y_n \in S(x_n)$ and $y \in S(x)$ such that $y_n \to y$ strongly.
Is this enough to conclude that $S$ has a fixed point? I.e. that there is a $z$ such that $z \in S(z)$? If $S$ were single valued, it is enough by Tikhonov's Fixed Point Theorem.
Is there something easy to apply in this case too?
Generally no. One usually also wants some regularity condition on the sets $S(x)$. The standard requirement is for $S(x)$ to be convex for all $x$, see here, for example.$^1$ (Note: this is vacuously satisfied when considering functions, hence why such a condition is not stated for Tikhonov's fixed point theorem.)
As a simple counter example of what can go wrong without such a hypothesis, let $X = Y = \mathbb{R}$, and let $$ S(x) = \begin{cases} -1 & \textrm{ if } x \ge 0 \\ 1 & \textrm{ if } x \le 0. \end{cases} $$ $S$ clearly takes the unit ball about the origin to itself, is upper-hemicontinuous, and fixed point free.
$^1$ Convexity can be greatly relaxed. The search terms to look up are 'acyclicity', 'Eilenberg-Montgomery fixed point theorem', and 'Vietoris-Begle theorem'. Gorniewicz' Topological Fixed Point Theory of Multivalued Mappings is a comprehensive treatment of the subject that may interest you.