Conditional expectation minimises mean squared error

Let $X, Y$ be square-integrable. Prove that

$$\E\big[(Y - g(X))^2\big] \ \ge\ \E\big[(Y - \E[Y\mid X])^2\big]$$

for every measurable $g$ with $\E[g(X)^2]<\infty$, with equality iff $g(X)=\E[Y\mid X]$ almost surely. Conclude that the Bayes-optimal predictor for squared loss is the regression function.