I am trying to show that:
$$\mathbb{V}[X+Y] = \mathbb{V}[X]+2\text{Cov}[XY]+\mathbb{V}[Y]$$
We have that:
$$\mathbb{V}[X+Y] = \mathbb{E}[(X+Y)^2]-\mathbb{E}[X+Y]^2$$
From the definition of covariance:
$$\text{Cov}[XY] = \mathbb{E}[X-\mathbb{E}[X]]\mathbb{E}[Y-\mathbb{E}[Y]]$$
First question: How come the above makes any sense at all? i.e. we know that expectation is idempotent, so: $\mathbb{E}[\mathbb{E}[X]]=\mathbb{E}[X]$, this mans that $\mathbb{E}[X-\mathbb{E}[X]] = \mathbb{E}[X]-\mathbb{E}[X]=0$...
I proceed to expand the second line of maths:
$$\mathbb{V}[X+Y] = \mathbb{E}[X^2] + 2\mathbb{E}[XY]+\mathbb{E}[Y^2] -(\mathbb{E}[X]^2 + 2\mathbb{E}[X]\mathbb{E}[Y]+\mathbb{E}[Y]^2)$$
Second question: How does $\mathbb{E}[X-\mathbb{E}[X]]\mathbb{E}[Y-\mathbb{E}[Y]] = \mathbb{E}[XY]-\mathbb{E}[X]\mathbb{E}[Y]$?
It doesn't because you have the definition wrong. It an expectation of a product not the product of expectations
$$\begin{align}\mathsf{Cov}[XY] = & ~ \mathsf{E}\big[(X-\mathsf{E}[X])(Y-\mathsf{E}[Y])\big] \\ = & ~ \mathsf E\big(XY-X~\mathsf E(Y)-\mathsf E(X)~Y+\mathsf E(X)~\mathsf E(Y)\big) \\ = & ~ \mathsf E(XY)-\mathsf E(X~\mathsf E(Y))-\mathsf E(\mathsf E(X)~Y)+\mathsf E(\mathsf E(X)~\mathsf E(Y)) \\ = & ~ \mathsf E(XY)-\mathsf E(X)~\mathsf E(Y)-\mathsf E(X)~\mathsf E(Y)+\mathsf E(X)~\mathsf E(Y) \\ = & ~ \mathsf E(XY)-\mathsf E(X)~\mathsf E(Y) \end{align}$$