Let $X$ be a random variable on the probability space $(\Omega, \mathcal A, P)$. Suppose $X$ is square-integrable and $\mathcal H$ is a sub $\sigma$-algebra of $\mathcal A$. Then $\mathbb E[X\mid \mathcal H]$ is square-integrable and for all square-integrable and $\mathcal H$-measurable $Z$ it holds that
$$\mathbb E[(X-\mathbb E[X\mid \mathcal H])^2] \le \mathbb E[(X-Z)^2]$$
To show that $\mathbb E[X\mid \mathcal H]$ is square-integrable I first note that $|\mathbb E[X\mid \mathcal H]|^2\le \mathbb E[|X|^2\mid \mathcal H]$ by the conditional Jensen's inequality. Now, taking expecations gives $$\mathbb E[|\mathbb E[X\mid\mathcal H]|^2]\le \mathbb E[\mathbb E[|X|^2\mid \mathcal H]] = \mathbb E[|X|^2] <\infty$$ So $\mathbb E[X\mid \mathcal H]$ is indeed square-integrable.
But how to do the second part?
Hint: For every $Z$ in $L^2(\mathcal H)$, $$X-Z=X-\mathbb E[X\mid \mathcal H]+\mathbb E[X\mid \mathcal H]-Z,$$ where $\mathbb E[X\mid \mathcal H]-Z$ is in $L^2(\mathcal H)$, and $X-\mathbb E[X\mid \mathcal H]$ is orthogonal to $L^2(\mathcal H)$. Thus, good ol'Pythagoras saves the day.