It is stated that $E[XY] = E[X]E[Y]$ only if $X$ and $Y$ are uncorrelated, thus $Cov(X,Y) = 0$.
The covariance of a random variable with itself is just it's variance, which is to say $Cov(X,X) = Var(X)$.
So it stands to reason that $E[X]E[X] \neq E[X^2]$, based on the covariance of $X$ being equal to $Var(X)$, which is probably not $0$.
However, on a homework question I've found it stated that: $E[(X-Y)]^2 = E[X^2] + E[Y^2] - 2E[XY]$, given that $X$ and $Y$ are independent variables.
It was further stated later in the question that $E[(X-Y)]^2 = E[X^2] - 2E[XY] + (Var(Y) + E(Y^2))$
Which makes me think it was not a mistake with parenthesis placement.
Should it not be:
$E[(X-Y)]^2 = (E[X]-E[Y])(E[X]-E[Y]) = E[X]^2 + E[Y]^2 - 2E[XY]$?
In general, $$(X-Y)^2 = X^2 + Y^2 - 2XY$$ Hence, by linearity of $\mathbb{E}$, $$\mathbb{E}[(X-Y)^2] = \mathbb{E}[X^2]+\mathbb{E}[Y^2]-2\mathbb{E}[XY]$$ Independence of $X$ and $Y$ is not required for the above. (Notice that I didn't use this assumption at all.)
Looking at the work you have, what you basically said was that $$\begin{align} \mathbb{E}[(X-Y)^2]&=\mathbb{E}[(X-Y)(X-Y)] \\ &= \mathbb{E}[X-Y]\cdot\mathbb{E}[X-Y]\quad\text{(not true!)} \end{align}$$ To split the expected values via multiplication, you must have independence of each multiplicative term. This isn't true, because $X-Y$ is obviously not independent of $X-Y$: observe that $$\text{Cov}(X-Y, X-Y)=\text{Var}[X-Y] = \text{Var}(X)+\text{Var}(Y)-2\cdot\text{Cov}(X, Y)$$ which - assuming $X$ and $Y$ are independent - as long as $\text{Var}(X)$ and $\text{Var}(Y)$ are not $0$, $\text{Cov}(X-Y, X-Y) \neq 0$, hence $X-Y$ is not independent of $X-Y$.