The derivative keeps all roots real

Idea. Count the roots of $P'$ that we can *see* and show that there are already $n-1=\deg P'$ of them.

Roots at the $r_i$. Write $P(x)=(x-r_i)^{m_i}Q(x)$ with $Q(r_i)\ne0$. Then $P'(x)=(x-r_i)^{m_i-1}\bigl(m_iQ(x)+(x-r_i)Q'(x)\bigr)$ and the second factor equals $m_iQ(r_i)\ne0$ at $r_i$. So $r_i$ is a root of $P'$ of multiplicity exactly $m_i-1$ (no root if $m_i=1$). Together these give $\sum_i(m_i-1)=n-k$ roots of $P'$.

Roots between the $r_i$. For $1\le i\le k-1$, $P(r_i)=P(r_{i+1})=0$, so by Rolle's theorem $P'$ vanishes at some point of the open interval $(r_i,r_{i+1})$. These $k-1$ points are distinct from each other and from all the $r_j$.

Conclusion. $P'$ has at least $(n-k)+(k-1)=n-1$ real roots counted with multiplicity. Since $\deg P'=n-1$, these are all its roots, and they are all real. $\blacksquare$

Example. $P(x)=x^3-3x+2=(x-1)^2(x+2)$: $P'(x)=3(x-1)(x+1)$, with the roots $1$ (from the double root) and $-1\in(-2,1)$ (from Rolle).

Lesson. Rolle's theorem counts roots: between two roots of $P$ there is a root of $P'$, and a multiple root of $P$ stays a root of $P'$. Count both and compare with the degree.