We want to show the following relationship:

$\mathbb E[X^2] = \mathbb E[(X - \mathbb E[X])^2] + \mathbb E[X]^2 \tag{1}$

If we expand the first expression on the right-hand side of Equation 1:

\begin{align} \mathbb E [(X - \mathbb E[X])^2] &= \mathbb E[X^2-2X\mathbb E[X]+\mathbb E [X]^2] \\ &= \mathbb E[X^2] - 2\mathbb E[X] \mathbb E [\mathbb E [X]] + \mathbb E [\mathbb E [X]^2]. \tag{2} \end{align}

If we note that $$\mathbb E[X]$$ is a constant, then $$\mathbb E[\mathbb E[X]]$$ is also a constant, namely $$\mathbb E[X]$$. Taking this into account Equation 2 simplifies to:

\begin{align} \mathbb E[(X - \mathbb E[X])^2] &= \mathbb E[X^2] - 2\mathbb E[X] \mathbb E[X] + \mathbb E[X]^2 \\ &= \mathbb E[X^2] - 2\mathbb E[X]^2 + \mathbb E[X]^2 \\ &= \mathbb E[X^2] - \mathbb E[X]^2 \tag{3} \end{align}

Plugging Equation 3 back into the right-hand side of Equation 1 gives

\begin{align} \mathbb E[(X - \mathbb E[X])^2] + \mathbb E[X]^2 &= \mathbb E[X^2] - \mathbb E[X]^2 + \mathbb E[X]^2 \\ &= \mathbb E[X^2] , \tag{4} \end{align}

thus giving the desired result.