Understanding the minimization of mean squared error function I try to minimize mean squared error function defined as:
$E\left[Y - f(X)\right]^2$
I summarized the minimization procedure from different online sources (e.g., URL 1 (p. 4), URL 2 (p. 8)) in the following lines.
First add and subtract $E[Y | X]$:
$E\left[\left\lbrace(Y - E[Y | X]) - (f(X) - E[Y|X])\right\rbrace^2\right]$
Expanding the quadratic yield:
$E\left[\left(Y - E[Y|X]\right)^2 + \left(f(X) - E[Y|X]\right)^2 - 2 \left(Y - E[Y|X]\right)\left(f(X) - E[Y|X]\right)\right]$
First term is not affected by the choice of $f(X)$; third term is $0$, so the whole expression is minimized if $f(X) = E(Y|X)$.
Question 1: I wonder what is the motivation to add and subtract $E[Y | X]$ in the first step of the procedure?
Question 2: How to explain in plain English why third term in quadratic is $0$?
 A: Concerning your first question, adding and subtracting is a trick in statistics which is often used to more easily work with certain expressions. By adding and subtracting you do not change your equation but it makes it possible to group certain terms to obtain the result more easily.
For your second question, to make this point more formally, we want to show the conditional expectation function (CEF) prediction property:
$$E(Y|X) = \text{arg min}_{f(X)} E[(Y-f(X))^2]$$
I guess that not stating that $f(X)$ is the minimization argument in the question caused confusion for some. The CEF also has the following decomposition property:
$$Y=E(Y|X) + \epsilon $$
where $\epsilon $ is a random variable such that $E(\epsilon|X) =0$ and $E(h(X)\epsilon)=0$.
In your last expression you have $(Y-E(Y|X)) = \epsilon$ and $(f(X)-E(Y|X)) = h(X)$ is a function of $X$. Then you use the previous property of $\epsilon$ to show that $-2E[h(X)\epsilon]=0$, hence the last expression is zero. This proof goes by using properties of the CEF rather than anything unnecessarily complicated - so it's plain English for most parts.
A: For your second question, you want to show that 
$$E\left[ (Y-E(Y|X)(f(X)-E(Y|X))\right]=0.$$
Now, if we look at the first term of the product, if we didn't have a conditional expectation, we would have
$$E(Y-E(Y))=E(Y)-E(Y)=0.$$
But by the Law of Total expectation, we know that
$$E(W)=E(E(W|Z)),$$
so you can actually write
$$E(Y-E(Y|X)) = E(E(Y-E(Y|X)|X)) = E(E(Y|X)-E(Y|X)) =E(0)=0.$$
To finish the proof, note that conditional on $X$, the second term is a constant, and therefore the expectation of the product is the product of the expectations:
$$E\left[ (Y-E(Y|X))(f(X)-E(Y|X))|X\right]=E\left[ (Y-E(Y|X))|X\right]\cdot E\left[(f(X)-E(Y|X))|X\right]$$
