Topology on the space of measurable functions
Not really. There are many compact metrizable topologies you can put on this space, but none that relate meaningfully to the structure of the problem.
Let's look first at the case $A_1=[0,1]$ and $A_2=\{0,1\}$. Consider the elevation function $e:A_1\times\Sigma\to\Delta(A_2)=[0,1]$ given by $e(a,\eta)=\eta(a)$. If you want the ultimate action choice of player $2$ to be continuous in player $1$'s action and the strategy choice of player $2$, you need $e$ to be continuous. In particular, $e$ should be measurable when you endow $\Sigma$ with the Borel $\sigma$-algebra. That never works:
It follows from the main result (Theorem D) in Robert Aumann's 1961 paper &quot;Borel structures for function spaces.&quot; that there is no Borel $\sigma$-algebra on $\Sigma$ that makes $e$ jointly measurable. For consumers,  I would recommend to read Aumann's 1963 article &quot;On choosing a function at random.&quot; instead.
To make this a bit more digestible, let $F$ be the set of measurable functions from the unit interval to itself. Let $e:[0,1]\times F\to [0,1]$ be given by $e(x,f)=f(x)$. If we endow $[0,1]$ with the Borel $\sigma$-algebra and $F$ with any (!) $\sigma$-algebra, $e$ will necessarily be nonmeasurble. Now, by Kuratowski's isomorphism theorem, all uncountable separable and complete metric spaces are isomorphic as measurable spaces. This will include $A_1$ (if uncountable) and $\Delta(A_2)$ (if $A_2$ has at least $2$ elements).