[proofplan]
We prove the comparison theorem by contradiction. Assuming that $u$ has no zero on $[c_1,c_2]$, we divide by $u$ and use the Picone square identity to compare the two Sturm-Liouville equations on the nodal interval of $v$. The endpoint conditions for $v$ remove boundary terms, and the coefficient inequalities force the integral of a nonnegative square to be at most $0$. Equality then makes $v/u$ constant, which contradicts the endpoint zeros of the nontrivial function $v$.
[/proofplan]
[step:Assume $u$ is positive on the nodal interval]
Suppose that $u$ has no zero in $[c_1,c_2]$. Since $u \in C^1([a,b];\mathbb{R})$, continuity gives either $u(x)>0$ for every $x \in [c_1,c_2]$ or $u(x)<0$ for every $x \in [c_1,c_2]$. Multiplying $u$ by $-1$ if necessary, assume $u(x)>0$ for every $x \in [c_1,c_2]$. The hypothesis that $v$ is not identically zero on $[c_1,c_2]$ will be used at the final contradiction.
[/step]
[step:Apply Picone's identity to $u$ and $v$]
Let $\mathcal{L}^1$ denote one-dimensional Lebesgue measure on $\mathbb{R}$. Define the comparison quotient $w: [c_1,c_2] \to \mathbb{R}$ by
\begin{align*}
w(x) = \frac{v(x)^2}{u(x)}.
\end{align*}
Because $u \in C^1([a,b];\mathbb{R})$, $v \in C^1([a,b];\mathbb{R})$, and $u>0$ on $[c_1,c_2]$, the quotient rule gives $w \in C^1([c_1,c_2];\mathbb{R})$. Since $v(c_1)=v(c_2)=0$, also $w(c_1)=w(c_2)=0$. Since $c_1,c_2 \in (a,b)$, there exists an open interval $I \subset (a,b)$ with $[c_1,c_2] \subset I$. The hypothesis $P_1u' \in C^1((a,b);\mathbb{R})$ therefore gives $P_1u' \in C^1(I;\mathbb{R})$, so multiplying the equation for $u$ by $w$ and applying one-dimensional integration by parts on $[c_1,c_2]$ gives
\begin{align*}
0
&= \int_{c_1}^{c_2} \left(P_1(x)u'(x)w'(x)+Q_1(x)u(x)w(x)\right)\,d\mathcal{L}^1(x).
\end{align*}
Since $u(x)w(x)=v(x)^2$, this yields
\begin{align*}
\int_{c_1}^{c_2} P_1(x)\left(v'(x)-\frac{v(x)u'(x)}{u(x)}\right)^2\,d\mathcal{L}^1(x)
= \int_{c_1}^{c_2}\left(P_1(x)v'(x)^2+Q_1(x)v(x)^2\right)\,d\mathcal{L}^1(x).
\end{align*}
[guided]
The purpose of introducing $w=v^2/u$ is that it converts the equation for $u$ into a perfect square. This is the Picone square computation; we prove the needed identity directly, so no external theorem is being invoked without hypotheses. Throughout the integrals below, $\mathcal{L}^1$ denotes one-dimensional Lebesgue measure on $\mathbb{R}$.
Define the comparison quotient $w: [c_1,c_2] \to \mathbb{R}$ by
\begin{align*}
w(x)=\frac{v(x)^2}{u(x)}.
\end{align*}
Because $u \in C^1([a,b];\mathbb{R})$, $v \in C^1([a,b];\mathbb{R})$, and $u(x)>0$ on $[c_1,c_2]$, the quotient rule gives $w \in C^1([c_1,c_2];\mathbb{R})$. The endpoint zeros of $v$ imply
\begin{align*}
w(c_1)=w(c_2)=0.
\end{align*}
Since $c_1,c_2 \in (a,b)$, choose an open interval $I \subset (a,b)$ such that $[c_1,c_2] \subset I$. The hypothesis $P_1u' \in C^1((a,b);\mathbb{R})$ implies $P_1u' \in C^1(I;\mathbb{R})$, so the product $(P_1u')'w$ is continuous on $[c_1,c_2]$ and integration by parts is valid on $[c_1,c_2]$. Multiplying
\begin{align*}
-(P_1u')'+Q_1u=0
\end{align*}
by $w$ and integrating over $[c_1,c_2]$ with respect to $\mathcal{L}^1$ gives
\begin{align*}
0
&=\int_{c_1}^{c_2}\left(P_1(x)u'(x)w'(x)+Q_1(x)u(x)w(x)\right)\,d\mathcal{L}^1(x).
\end{align*}
Here the boundary term is $[-P_1u'w]_{c_1}^{c_2}$, and it vanishes because $w(c_1)=w(c_2)=0$.
Now compute the derivative of the quotient:
\begin{align*}
w'(x)=\frac{2v(x)v'(x)u(x)-v(x)^2u'(x)}{u(x)^2}.
\end{align*}
Substituting this expression into the integrand gives
\begin{align*}
P_1(x)v'(x)^2-P_1(x)u'(x)w'(x)
&=P_1(x)\left(v'(x)^2-2\frac{v(x)v'(x)u'(x)}{u(x)}+\frac{v(x)^2u'(x)^2}{u(x)^2}\right) \\
&=P_1(x)\left(v'(x)-\frac{v(x)u'(x)}{u(x)}\right)^2.
\end{align*}
Since $u(x)w(x)=v(x)^2$, rearranging the integrated equation for $u$ yields
\begin{align*}
\int_{c_1}^{c_2} P_1(x)\left(v'(x)-\frac{v(x)u'(x)}{u(x)}\right)^2\,d\mathcal{L}^1(x)
= \int_{c_1}^{c_2}\left(P_1(x)v'(x)^2+Q_1(x)v(x)^2\right)\,d\mathcal{L}^1(x).
\end{align*}
[/guided]
[/step]
[step:Use the equation for $v$ to compare the two energies]
Since $v \in C^1([a,b];\mathbb{R})$ and $c_1,c_2 \in (a,b)$, choose an open interval $J \subset (a,b)$ such that $[c_1,c_2] \subset J$. The hypothesis $P_2v' \in C^1((a,b);\mathbb{R})$ gives $P_2v' \in C^1(J;\mathbb{R})$, so multiplying the equation for $v$ by $v$ and applying one-dimensional integration by parts over $[c_1,c_2]$ with respect to $\mathcal{L}^1$ is valid. The boundary term $[-P_2v'v]_{c_1}^{c_2}$ vanishes because $v(c_1)=v(c_2)=0$, so
\begin{align*}
0
&=\int_{c_1}^{c_2}\left(P_2(x)v'(x)^2+Q_2(x)v(x)^2\right)\,d\mathcal{L}^1(x).
\end{align*}
Hence
\begin{align*}
\int_{c_1}^{c_2} P_1(x)\left(v'(x)-\frac{v(x)u'(x)}{u(x)}\right)^2\,d\mathcal{L}^1(x)
&=\int_{c_1}^{c_2}\left((P_1(x)-P_2(x))v'(x)^2+(Q_1(x)-Q_2(x))v(x)^2\right)\,d\mathcal{L}^1(x) \\
&\leq 0,
\end{align*}
because $P_1\leq P_2$ and $Q_1\leq Q_2$ on $[c_1,c_2]$.
[/step]
[step:Show the equality case contradicts the endpoint zeros of $v$]
The left-hand side is the integral of the nonnegative continuous function
\begin{align*}
x \mapsto P_1(x)\left(v'(x)-\frac{v(x)u'(x)}{u(x)}\right)^2.
\end{align*}
The previous step shows that this integral is at most $0$. Hence the integral is $0$, and nonnegativity plus continuity imply
\begin{align*}
P_1(x)\left(v'(x)-\frac{v(x)u'(x)}{u(x)}\right)^2=0
\end{align*}
for every $x\in[c_1,c_2]$. Since $P_1(x)>0$ on $[c_1,c_2]$, we obtain
\begin{align*}
v'(x)-\frac{v(x)u'(x)}{u(x)}=0
\end{align*}
for every $x\in[c_1,c_2]$. Because $u(x)>0$ on $[c_1,c_2]$, the quotient map $r:[c_1,c_2]\to\mathbb{R}$ defined by $r(x)=v(x)/u(x)$ belongs to $C^1([c_1,c_2];\mathbb{R})$, and
\begin{align*}
r'(x)=\frac{u(x)v'(x)-v(x)u'(x)}{u(x)^2}=0
\end{align*}
for every $x\in[c_1,c_2]$. Therefore $r$ is constant on $[c_1,c_2]$; write $r(x)=\lambda$ for a constant $\lambda\in\mathbb{R}$. Since $v(c_1)=0$ and $u(c_1)>0$, we have $\lambda=0$. Hence $v(x)=0$ for every $x\in[c_1,c_2]$, contradicting the hypothesis that $v$ is not identically zero on $[c_1,c_2]$. The assumption that $u$ has no zero on $[c_1,c_2]$ is false, so $u$ has a zero in $[c_1,c_2]$.
[guided]
The comparison step has placed us in a rigid equality situation. We know
\begin{align*}
\int_{c_1}^{c_2} P_1(x)\left(v'(x)-\frac{v(x)u'(x)}{u(x)}\right)^2\,d\mathcal{L}^1(x) \leq 0.
\end{align*}
The integrand is nonnegative because $P_1(x)>0$ and the remaining factor is a square. It is also continuous on $[c_1,c_2]$: the functions $u$ and $v$ are $C^1$, and $u(x)>0$ on the whole interval, so division by $u$ is continuous. Therefore a continuous nonnegative function with integral $0$ must vanish at every point, and we get
\begin{align*}
P_1(x)\left(v'(x)-\frac{v(x)u'(x)}{u(x)}\right)^2=0
\end{align*}
for every $x\in[c_1,c_2]$.
Since $P_1(x)>0$, the square itself vanishes:
\begin{align*}
v'(x)-\frac{v(x)u'(x)}{u(x)}=0.
\end{align*}
This equation is exactly the statement that the quotient $v/u$ has derivative zero. Indeed, define $r:[c_1,c_2]\to\mathbb{R}$ by $r(x)=v(x)/u(x)$. The quotient is well-defined and belongs to $C^1([c_1,c_2];\mathbb{R})$ because $u>0$ and $u,v\in C^1([a,b];\mathbb{R})$. The quotient rule gives
\begin{align*}
r'(x)=\frac{u(x)v'(x)-v(x)u'(x)}{u(x)^2}=0
\end{align*}
for every $x\in[c_1,c_2]$.
Thus $r$ is constant on $[c_1,c_2]$; let $\lambda\in\mathbb{R}$ denote this constant. Then $v=\lambda u$ on $[c_1,c_2]$. Now the endpoint zero at $c_1$ forces the constant to be zero, because
\begin{align*}
0=v(c_1)=\lambda u(c_1)
\end{align*}
and $u(c_1)>0$. Hence $\lambda=0$, so $v=0$ on all of $[c_1,c_2]$. This contradicts the hypothesis that $v$ is not identically zero on $[c_1,c_2]$. Therefore the contradiction assumption was impossible, and $u$ must have a zero in $[c_1,c_2]$.
[/guided]
[/step]
