Let's start with the case where $P$ is of degree 2, with no real roots. In this case, we can write $P(x) = a x^2 + b x + c$ with $a > 0$ and $b^2 - 4ac < 0$.

Now let's consider the case where $P$ has only real roots. The degree of $P$ is obviously even, otherwise $P$ would tend to $-\infty$ as $x \to +\infty$ or $-\infty$. We work by induction on $\operatorname{deg}(P)$. If $\operatorname{deg}(P) = 0$, $P$ is a positive constant, hence it is the square of a real number. If the result is true for a polynomial with all real roots of degree $2n$ and $P$ is of degree $2n+2$ with all real roots, let $a$ be a root of $P$. If $a$ is a simple root of $P$, then $P$ changes sign at $a$, which is absurd since $P(x) \geq 0$ for all $x$. Thus, $a$ is a multiple root of $P$, and we can write $P(x) = (x-a)^2 Q(x)$, where $Q$ is of degree $2n$, with all real roots, and always positive. By induction, there exist two polynomials $A$ and $B$ such that $Q = A^2 + B^2$, then $P(x) = ((x-a) A(x))^2 + ((x-a) B(x))^2$, which concludes the induction.

Finally, we handle the general case. Since $P$ has real coefficients, it can be written as a product of a polynomial with all real roots and polynomials of degree 2 with no real roots. Each of these polynomials can be written as a sum of two squares as we have shown before. To finish, it remains to see that if $A, B, C, D$ are four polynomials, we have

$$\left(A^2 + B^2\right)\left(C^2 + D^2\right) = (A C - B D)^2 + (A D + B C)^2$$

We can recover this formula by thinking of the equality in the complex numbers

$$(a + i b)(c + i d) = (a c - b d) + i(a d + b c)$$

which gives, by taking the norm and squaring

$$\left(a^2 + b^2\right)\left(c^2 + d^2\right) = (a c - b d)^2 + (a d + b c)^2$$