Proof of $E(E(X|Y))$ Why is 
$$
E(E(X|Y)) = \int_{\infty}^{\infty} E(X|Y)f_2(y)dy
$$
So, another way of putting it, is why is this, given 
$$
E(E(X|Y)) = E(h(y))
$$
Why do they take the above function to be true? And a fuction of y, and not x?

The argument is straightforward when everything is phrased correctly. I'll use capitals $X,Y$ for random variables, and lowercase $x,y$ for the values they can take.
Let $h(y) = E(X|Y=y)$ be the expected value of $X$ given that $Y=y$. Obviously normal expectations $E(X)$ are just numbers, not functions, meaning things like "the average of $X$". This is just saying "What is the average of $X$ when $Y=y$?" The only thing this can depend on is $y$.

Let's spell everything out in gory detail.
Now given that $X,Y$ have the joint distribution function $f_{X,Y}(x,y)$, we find
$$\begin{align}h(y) = E(X|Y=y) & = \int_{-\infty}^\infty x f_{X|Y}(x|y)\,\mathrm dx \\
& = \int_{-\infty}^\infty x \frac{f_{X,Y}(x,y)}{f_Y(y)}\,\mathrm dx \\
& = \frac{1}{f_Y(y)}\int_{-\infty}^\infty x f_{X,Y}(x,y)\,\mathrm dx
\end{align}$$
Therefore, $E(h(Y))$ is given by
$$
\begin{align}\int f_Y(y)h(y)\mathrm d y & = \int \mathrm d y f_Y(y) \frac{1}{f_Y(y)}\int x f_{X,Y}(x,y)\,\mathrm dx \\
& = \int \mathrm d y \int x f_{X,Y}(x,y)\,\mathrm dx \\
& = \int x \left[ \int \mathrm d y f_{X,Y}(x,y) \right] \,\mathrm dx \\
& = \int x f_X(x) \,\mathrm dx \\
& = E(X)
\end{align}$$
