Let $h:\Bbb R\mapsto \Bbb R$ be a bounded function. By defining $\tilde h:\Bbb R^2\mapsto \Bbb R$ as $\tilde h(x,y)=h(x)~\forall y\in\Bbb R$ we have: $$\begin{align}\Bbb E(\Bbb E(Z\mid X)h(X))&=\Bbb E(Zh(X)) \\ & =\Bbb E(Z\tilde h(X,Y)) \\ & =\Bbb E(\Bbb E(Z\mid X,Y)\tilde h(X,Y)) \\ & =\Bbb E(\Bbb E(Z\mid X,Y)h(X)) \end{align}$$
Which implies $ \Bbb E(Z\mid X)\Bbb E(h(X))=\Bbb E(Z\mid X,Y)\Bbb E(h(X))\implies \Bbb E(Z\mid X)=\Bbb E(Z\mid X,Y)$
From this we can get $\Bbb E(\Bbb E(Z\mid X)\mid X)=\Bbb E(\Bbb E(Z\mid X,Y)\mid X)$ but is the left hand side equal to $\Bbb E(Z\mid X)$, and why? (That would conclude the proof)
I do not follow where you say "which implies"... As I understand you use the defintion: $Y$ is a conditional expectation given $X$ iff for all bounded measurable functions $h$ the following equality holds: \begin{align} \mathbb E(Yh(X))=\mathbb E(Xh(X)) \end{align} Is this right?
In your specific case you want to pove \begin{align} \mathbb E(\mathbb E(Z\mid X)h(X))=\mathbb E(\mathbb E(\mathbb E(Z\mid X,Y)\mid X)h(X)) \end{align} You are almost there, because what you have showed is \begin{align} \mathbb E(\mathbb E(Z\mid X)h(X))=\mathbb E(\mathbb E(Z\mid X,Y)h(X)) \end{align} We go one step further and use the "pull-out what is known" property and the law of total expectation \begin{align} \mathbb E(\mathbb E(Z\mid X)h(X))&=\mathbb E(\mathbb E(Z\mid X,Y)h(X))\\ &=\mathbb E(\mathbb E(\mathbb E(Z\mid X,Y)h(X)\mid X))\\ &=\mathbb E(\mathbb E(\mathbb E(Z\mid X,Y)\mid X)h(X)) \end{align} If you do not know the "pull-out what is known" property, that is $\mathbb E(Yg(X)\mid X)=g(X)\mathbb E(Y\mid X)$. How to prove it? Well \begin{align} \mathbb E(\mathbb E(Yg(X)\mid X)h(X))=\mathbb E(Yg(X)h(X))=\mathbb E(\mathbb E(Y\mid X)h(X)g(X)) \end{align} Conclude..