$E(E(Z\mid X,Y)\mid X)=E(Z\mid X)$
The conditional expectation $E(Y\mid X)$ is defined as the (almost surely unique) function $\phi(X)$ such that $E(\phi(X)h(X))=E(Yh(X))$ for any bounded function $h$.
Now to prove the property, why is it enough to show that for any bounded function $h$ we have $$E(E(Z\mid X) h(X))=E(E(Z\mid X,Y)h(X))$$
It seems weird that but I would be tempted to say that we need to verify that $$E(E(Z\mid X) h(X))=E(E(E(Z\mid X,Y)\mid X)h(X))$$
In general, in order to prove that $\mathbb E \left[Z_1\mid X\right]=\mathbb E \left[Z_2\mid X\right]$, it suffices, with the mentioned definition of conditional expectation, to show that that for each bounded function $h$, $$\tag{*} \mathbb E\left[Z_1h(X)\right]=\mathbb E\left[Z_2h(X)\right]. $$ Indeed, if $\phi_1$ and $\phi_2$ are such that $\mathbb E \left[Z_i\mid X\right]=\phi_i(X)$, $i=1,2$, equality (*) implies that for each bounded function $h$, $$ \mathbb E\left[\phi_1(X)h(X)\right]=\mathbb E\left[\phi_2(X)h(X)\right] $$ or equivalenty, that $$ \mathbb E\left[(\phi_1-\phi_2)(X)h(X)\right]=0. $$ Use this with $h\colon x\mapsto \operatorname{sgn}(\phi_1(x)-\phi_2(x))$.