Prove that, if $y''+w(x)y=0,$ then $y$ has infinitely many zeroes

I didn't know of the Sturm-Picone comparison theorem before, but after having looked it up, it totally makes sense to use it. In the notation of the Wikipedia article you will have $p_1=p_2=1,$ $q_1=1$ and $q_2=q.$ The proof can hardly be simpler.

Here is a proof that does not rely on the Sturm–Picone comparison theorem. The most important thing about $w(x)$ is that it is bounded below by some $\epsilon>0$ (in your case $\epsilon=1$). Now, suppose that $y$ has a finite number of zeros and let $x_0$ be the largest zero plus $1$. Then for $x\geq x_0$, $y<0$ or $y>0$.

Let us consider the first case (the second follows in a similar manner). Since $y<0$ and $w(x)>\epsilon>0$, we know $x\geq x_0$ implies $y''>0$. However, we can say something further: that $y''>\delta>0$ (it is bounded away from zero by some $\delta$). To see this, note that for $z>x\geq x_0$ by the mean value theorem for integrals there exists $c\in (x,z)$ such that

$$\frac{1}{z-x}\int_x^zy'(t)dt=y(c)<0$$

Now, if $y'(x_1)>0$ for some $x_1>x_0$, let $\rho>0$ be defined such that $y'(x)\geq 0$ for $x\in [x_1-\rho,x_1+\rho]$. We can then restrict the interval of the integral such that there exists $c\in (x_1-\rho,x_1+\rho)$ such that

$$0>y(c)=\frac{1}{2\rho}\int_{x_1-\rho}^{x_1+\rho}y'(t)dt\geq \frac{1}{2\rho}\int_{x_1-\rho}^{x_1+\rho}0dt=0$$

a contradiction. Thus, $y'(x)\leq 0$ for all $x>x_0$. This implies $y(x)$ is decreasing on $[x_0,\infty)$ and therefore

$$y''(x)=-w(x)y(x)=w(x)|y(x)|>\epsilon |y(x_0)|>0$$

We now come to the conclusion of our proof. Using the fundamental theorem of calculus for $x>x_0$ we can write

$$y(x)=\int_{x_0}^x\left[ \int_{x_0}^zy''(t)dt+y'(x_0)\right]dz+y(x_0)$$

$$>\int_{x_0}^x\left[ \int_{x_0}^z\epsilon |y(x_0)|dt+y'(x_0)\right]dz+y(x_0)$$

$$=\frac{\epsilon |y(x_0)|}{2} x^2+(y'(x_0)-\epsilon|y(x_0)|)x+\frac{{x_0}^2\epsilon|y(x_0)|}{2}-x_0y'(x_0)+y(x_0)$$

Since this is a quadratic whose leading term is positive, we conclude $y(x)$ goes to infinity, a contradiction. As the case where $y>0$ is worked in the same manner (except with a sign change), we conclude $w(x)>\epsilon>0$ implies $y(x)$ has an infinite number of zeros.