Solution:

Since $P(x) \geq 0$ for all $x$, it can have no real roots except double roots, so we can write it as a product
$$P(x) = c \prod_{k=1}^{n} \left(x^{2} + p_{k} x + q_{k}\right)$$
of quadratics with nonpositive discriminant, i.e. $p_{k}^{2} - 4q_{k} \leq 0$. But then completing the square in each quadratic lets us write it as a sum of two squares of polynomials
$$x^{2} + p_{k} x + q_{k} = \left(x + \frac{p_{k}}{2}\right)^{2} + \left(\frac{\sqrt{4q_{k} - p_{k}^{2}}}{2}\right)^{2}$$
Thus, $P(x)$ is a product of sums of two squares. But now we note that for any polynomials $A(x), B(x), C(x), D(x)$ with real coefficients,
$$\left(A^{2}(x) + B^{2}(x)\right)\left(C^{2}(x) + D^{2}(x)\right) = (A(x)C(x) + B(x)D(x))^{2} + (A(x)D(x) - B(x)C(x))^{2}$$
i.e., a product of two sums of squares of polynomials is also a sum of two squares (Lagrange's identity for polynomials). Inductively applying this to the factors in $P(x)$ shows that $P(x)$ is also a sum of two squares.