If $W$ admits a periodic point, then, of course, $S(x)$ is constant along its orbit. But, on the contrary, whatever nice and well-structured $A$ and $W$ you have, if $W$ has no periodic points, there is always the bad map $S(x):=$ the negative invariant set generated by $x$, that is $cup_{kinmathbb{N}}W^{-k}(x),$ and this map fails to satisfy the thesis while $S(x)subset S(W(x))$ holds for all $x$.
This is to convince you that some assumption on the map $S(x)$ is in order. Here, a mild and natural assumption, to be coupled with the compactness of $Aneqemptyset$, is weak upper semicontinuity, that is,
$$ S:Ato2^X$$
is continuous w.r.to the product topology on $2^X$ where the two-point space $2:={0,1}$ is endowed with the left-order topology (whose only proper open subset is ${0}$). As a consequence, the continuous image of $A$ is a compact subset of $2^X$, therefore it has a maximal element with respect to inclusion. Due to your assumption on $W$, the equality necessarily holds for any maximal set $S(x),$ proving your thesis (incidentally, note that no further assumption on $W$ is needed).
Rmk 1. The upper semicontinuity of $S$ introduced above may be equivalently stated as:
$operatorname{graph(S)}:={(a,x)in Atimes X\ : ain S(x)}$ is closed in $Atimes X$, where $X$ has the discrete topology;
$S^* :Xto 2^A$ is a closed map, that is, for any $ain A$ the set $S^* (x):={ain A\,:\, xin S(a) }$ is a closed subset of $A$;
for any $ain A$ (denoting $mathcal{N}_ a$ the family of nbd's of $a$), there holds
$$limsup_{bto a} S(b) := cup_ {Uinmathcal{N}_ a} cap_ {bin U} S(b) subset S(a).$$
Rmk 2. The fact that a compact subset $K$ of $2^X$, where $2$ has the left-order topology, admits a maximal element, is, of course, a consequence of the Zorn lemma. Indeed, if $Gamma$ is an infinite chain in $K$, it has a limit point in $K$, which turns out to be an upper bound of $Gamma$.
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