I know the following.
$$Cov(X, E(Y|X)) = Cov(X,Y)$$
This implies that
$$Cov(Y, E(Y|X)) = Cov(Y,Y) = Var(Y)$$
And therefore
$Var(Y - E(Y|X)) = Var(Y) + Var(E(Y|X)) - 2Cov(Y, E(Y|X)) = Var(Y) + Var(E(Y|X)) - 2Var(Y) = Var(E(Y|X)) - Var(Y)$
And since $$Var(Y) = E(Var(Y|X)) + Var(E(Y|X))$$
I get that
$$Var(Y - E(Y|X)) = -E(Var(Y|X))$$
I cannot get rid of that minus sign. Any tips on where I'm going wrong?
On
Your first statement does hold, since \begin{align*} &Cov(X,\mathbb{E}(Y|X)) = \mathbb{E}[X\mathbb{E}[Y|X]] - \mathbb{E}[X]\mathbb{E}[ \mathbb{E}[Y|X]] = \mathbb{E}[\mathbb{E}[YX|X]] - \mathbb{E}[X]\mathbb{E}[Y] \\ &= \mathbb{E}[XY] - \mathbb{E}[X]\mathbb{E}[Y] = Cov(X,Y) \end{align*} where we use the Tower property for the second and last step. However, this does only imply that: \begin{align*} Cov(Y, \mathbb{E}(Y|Y)) = Cov(Y,Y), \end{align*} since you have set $Y=X$. This does not give any information about $Cov(Y,\mathbb{E}(Y|X))$.
The first equation does not imply the second equation, take for instance the case where $X$ and $Y$ are independent, then $$Cov(Y,E[Y|X])=Cov(Y,E[Y])=0$$, which is usually not the same as $Var(Y)$.
We can however solve it by using that $$Var(Y-E[Y|X]) =Var(E\big[ Y-E[Y|X]\: |X\big])+E[Var(Y-E[Y|X] \:| X)]$$ and realize that $E\big[ Y-E[Y|X]\: |X\big] = E[Y|X]-E[Y|X] = 0$, which means that $$Var(Y-E[Y|X]) = E[Var(Y-E[Y|X] \:| X)]$$ and from here, we are done if we can argue that $Var(Y-E[Y|X] \:| X)=Var(Y|X)$. I leave the last computation to you (simply apply the definition of the conditional variance).