Answer
See explanation
Work Step by Step
We will start on the right-hand side and show that it is equal to the left-hand side.
$\begin{align*}
E_X[E(Y\vert x)] &= \int_{-\infty}^\infty E(Y\vert x)\cdot f_X(x)\ dx \qquad \text{[ since}\ E_X[g(x)] = \int_{\mathbb{R}} g(x)f_X(x)\ dx\ ]\\
&= \int_{-\infty}^\infty \overbrace{\left( \int_{-\infty}^\infty y\cdot f_{Y\vert X=x}(y\vert x)\ dy\right)}^{E(Y\vert X=x)}\cdot f_X(x)\ dx \\
&= \int_{-\infty}^\infty \left( \int_{-\infty}^\infty y\cdot \overbrace{\frac{f_{X,Y}(x,y)}{f_X(x)}}^{f_{Y\vert X=x}(y\vert x)}\ dy\right)\cdot f_X(x)\ dx \\
&= \int_{-\infty}^\infty \left( \int_{-\infty}^\infty y\cdot\frac{f_{X,Y}(x,y)}{f_X(x)}\cdot f_X(x)\ dy\right)\ dx \\
& \qquad\qquad\qquad\qquad\qquad \text{[ since}\ f_X(x)\ \text{isn't a function of}\ y\ ]\\
&= \int_{-\infty}^\infty \left( \int_{-\infty}^\infty y\cdot f_{X,Y}(x,y)\ dy\right)\ dx \\
&= \int_{-\infty}^\infty \left( \int_{-\infty}^\infty y\cdot f_{X,Y}(x,y)\ dx\right)\ dy \qquad \text{[ switch order of integration ]}\\
&= \int_{-\infty}^\infty y\cdot \underbrace{\left( \int_{-\infty}^\infty f_{X,Y}(x,y)\ dx\right)}_{f_Y(y)}\ dy \\
&= \int_{-\infty}^\infty y\cdot f_Y(y)\ dy \\
E_X[E(Y\vert x)] &= E[Y] \\
&\text{Q.E.D}
\end{align*}$
