If $E(Y\mid X)$ is constant then $X, Y$ are uncorrelated. Last minute studying please tell me how to:

Prove that if the expected conditional expected value of the random variable $X$ given the random variable $Y$ - denoted by $E(X\mid Y)$ - is constant then $X, Y$ are uncorrelated. 

 A: If ${\rm E}[X\mid Y]$ is constant, then it must necessarily be equal to its mean ${\rm E}[{\rm E}[X\mid Y]]={\rm E}[X]$. Now use that
$$
{\rm E}[XY]={\rm E}[{\rm E}[XY\mid Y]]
$$
to conclude that ${\rm E}[XY]={\rm E}[X]{\rm E}[Y]$ which is equivalent to saying that the covariance is zero.
A: What you need is the double expectation formula:
$$
   \DeclareMathOperator{\E}{E}
    \E(X) = \E \E (X|Y) 
$$
In the double expectation, the inner expectation is a function of $Y$, that is, a random variable, and the outer expectation then is then taken over the distribution of $Y$. Assume $\E X = \E Y = 0$ (we can do that without loss of geneality, since else, just subtract first the expectation), and then $\E(X | Y)=s$, a known, constant real number. Now calculate
$$
\E XY = \E (\E (XY|Y)) = \E ( Y \E(X|Y)) = \E Ys = s\E Y =0.
$$
A: For X and Y to be uncorrelated we need to show $\rho(X, Y) = 0$
We know, $\rho(X, Y) = \frac{Cov(X,Y)}{\sigma_X\sigma_Y} = \frac{E(XY) - E(X)E(Y)}{\sigma_X\sigma_Y}$
For $\rho(X, Y)$ to be zero, the numerator has to be zero. Therefore it will suffice to show that $E(XY) = E(X)E(Y)$
From the question we know, $E(X|Y) = c$
We can calculate $E(X)$ using rule of Iterated Expectations,
$\begin{align} 
E(X) & = E[E(X|Y)] \\ 
& = E(c) \\
& = c
\end{align}$
Now, again using rule of Iterated Expectations on $E(XY)$
$\begin{align} 
E(XY) & = E[E[XY|Y]] \\
& = E[Y E(X|Y)]  &\text{;$E(XY|Y)=Y E(X|Y)$ when X and Y are conditionally independent} \\
& = E[Y c] \\
& = c E[Y] \\
& = E(X) E[Y] 
\end{align}$
Therefore,
$\rho(X, Y) = \frac{Cov(X,Y)}{\sigma_X\sigma_Y} = \frac{E(XY) - E(X)E(Y)}{\sigma_X\sigma_Y} = \frac{E(X)E(Y) - E(X)E(Y)}{\sigma_X\sigma_Y} = \frac{0}{\sigma_X\sigma_Y} = 0$

Proof of $E(XY|Y)=Y E(X|Y)$ when X and Y are conditionally independent.
We know if $X_1,...X_n$ be independent random variable, Then
$E(\prod_{i=1}^{n} X_i) = \prod_i E(X_i)$
Now, 
$\begin{align} 
E(XY|Y) & = E(X|Y) E(Y|Y)\\
& = E(X|Y) Y
\end{align}$
