What is $E[E[X|Y]|Y]$? What is $E[E[X|Y]|Y]$. Constant or random variable?
We know that $E[X|Y]$ is a random variable that dou should get epends on Y. Then the last given Y in $E[E[X|Y]|Y]$ should tell you what that Y is, so $E[E[X|Y]|Y]$ is a constant right?
It like saying $E[X|X]$ that should be a constant.
 A: $\mathrm{E}(X\,|\,X)$ is not a constant, it is equal to $X$. Similarly, $E(E(X\,|\,Y)\,|\,Y)$ is equal to $E(X\,|\,Y)$. How you can explain this is depending on how your definition of conditional expectation is. Informally, $E(X\,|\,Y)$ is a random variable, defined for all outcomes of $Y$, that is equal to the expectation of $X$ given this outcome of $Y$ ($E(X|Y=a)$). Conditioning on $Y$ again is trivial, since that would be a function equal to $E(E(X\,|\,Y)|Y = a)$ for every outcome $a$ of $Y$, so that is equal to $E(X|Y=a)$ for all outcomes, equivalent to conditioning one time. 
EDIT The important parts from Wikipedia I list below. This is not a real explanation, but if you are into the theory already, you may know where this fits in. 


*

*Definition


Let $\displaystyle g:U\to \mathbb {R} ^{n}$ be a $\Sigma$ -measurable function such that, for every $\displaystyle \Sigma $-measurable function $\displaystyle f:U\to \mathbb {R} ^{n}$,
$$\displaystyle \int g(Y)f(Y)\,dP=\int Xf(Y)\,dP.$$
Then the random variable$\displaystyle g(Y)$, denoted as $\displaystyle \operatorname {E} (X\mid Y)$, is a conditional expectation of X given $\displaystyle Y$ .


*Why $\mathrm{E}(X|X) = X$?


It's this property:
(Stability) If $\displaystyle X$ is $\displaystyle {\mathcal {H}}$-measurable, then $\displaystyle E(X\mid {\mathcal {H}})=X$. 
A: There are multiple common definitions/characterizations of the conditional expectation. Personally the one I find most insightful is the least squares characterization, that
$$ \mathbb{E}[X | Y] = g(Y) $$
for  $g$ the solution to
$$ \min_{g} \mathbb{E}[(X - g(Y))^{2}]. $$
In other words, sort of doing a (non-linear) least squares regression of $X$ on $Y$ to solve
$$ X = g(Y) + \epsilon $$
for $\epsilon$ the mean-zero "residual" which is independent of $Y$.
Then we can see that, for any function $f$,
$$ \mathbb{E}[f(Y) | Y] = f(Y), $$
so
$$ \mathbb{E}[\mathbb{E}[X|Y] | Y] =\mathbb{E}[g(Y) | Y] = g(Y). $$
This also works for understanding why
$\mathbb{E}[X | X] = X$, and also why
$$ \mathbb{E}[X] = \mathbb{E}[X | 1] $$
is a constant (or equivalently a constant random variable).
However, I do agree with the commenters that probably the proof meant $\mathbb{E}[\mathbb{E}[X | Y]]$, which is a constant, and the extra "$ | Y$" was just a typo or something.
