Here is a simple and explicit counterexample:
For $k=0,1,\dots$, let
\begin{equation}
y(x):=
\left\{
\begin{aligned}
(-2)^k\sin\Big(\frac\pi2\frac{x-x_{2k}}{x_{2k+1}-x_{2k}}\Big)&\text{ if }x_{2k}\le x\le x_{2k+1}, \\
(-2)^k\sin\Big(\frac\pi2\frac{x_{2k+2}-x}{x_{2k+2}-x_{2k+1}}\Big)&\text{ if }x_{2k+1}\le x\le x_{2k+2},
\end{aligned}
\right.
\end{equation}
where
\begin{equation}
x_n:=\sum_{j=1}^n h_j,
\end{equation}
$h_j:=1$ if $j$ is even, and $h_j:=2$ if $j$ is odd.
For all $j=0,1,\dots$, let also
\begin{equation}
V(x):=\frac{\pi^2}{h_j^2}\text{ if }x_j\le x<x_{j+1}.
\end{equation}
Then $y\in C^1[0,\infty)$ and
\begin{equation}
y''(x)+V(x)y(x)=0
\end{equation}
for all $x\in[0,\infty)\setminus\{x_1,x_2,\dots\}$. However,
\begin{equation}
|y(x_{2k+1})|=|(-2)^k|\to\infty
\end{equation}
as $k\to\infty$. To complete the construction, it remains to smooth $y$ appropriately at the points $x_1,x_2,\dots$.
Here is the plot $\{(x,y(x))\colon0\le x\le x_{12}\}=\{(x,y(x))\colon0\le x\le18\}$:
This illustrates the idea of this counterexample: let the (variable) frequency $\sqrt{V(x)}$ be increasing in $x$ between any two consecutive zeroes of $y$. Then the ratio $r_k$ of the value of $|y'|$ at each zero $z_k$ of $y$ to the value of $|y'|$ at the immediately preceding zero $z_{k-1}$ of $y$ will be greater than $1$, which will result in increasing amplitudes. The increase of the amplitudes will be exactly geometric if the ratio $r_k$ is constant, as in the above counterexample.
Here is another implementation of the same idea. This implementation does not require an additional smoothing. Let
\begin{equation}
f(x):=(1 - x) x (1 + x - x^2 + 5 x^3 - 4 x^4).
\end{equation}
Note that $f>0$ on $(0,1)$,
\begin{equation}
f(0)=f(1)=0=f''(0)=f''(1),\ f'(0)=1, f'(1)=-2,
\end{equation}
so that $|f'(1)|>|f'(0)|>0$; so, the graph of $f$ over $[0,1]$ is skewed a bit to the right:
For $k=0,1,\dots$, let now
\begin{equation}
Y(x):=(-2)^k f(t-k)\text{ if }k\le x\le k+1.
\end{equation}
For all $k=0,1,\dots$, let also $V\colon[0,\infty)\to\mathbb R$ be periodic with period $1$ and such that
\begin{equation}
V(x)=\frac{12 \left(1-5x+10 x^2\right)}{1+x-x^2+5 x^3-4 x^4}
\text{ if }0\le x<1.
\end{equation}
Then $3<V\le36$ on $[0,\infty)$,
$Y\in C^2[0,\infty)$, and
\begin{equation}
Y''(x)+V(x)Y(x)=0
\end{equation}
for all $x\in[0,\infty)$. However,
\begin{equation}
|Y(k+1/2)|=f(1/2)|(-2)^k|\to\infty
\end{equation}
as $k\to\infty$.
Here is the plot $\{(x,Y(x))\colon0\le x\le6\}$:
As a "real-world" application, especially of the first of the above two counterexamples, consider a basic LC oscillator circuit, with zero/negligible resistance; see e.g. Wikipedia and/or Electronics Tutorials. When the frequency of an external force applied to the circuit is the same as (or close to) the own, resonant frequency $\omega_0=1/\sqrt{LC}$ of the circuit, resonance can occur, with the oscillation amplitudes increasing many/(infinitely many) times. But this kind of resonance is caused by the external force and thus requires work, that is, expending energy.
Let us now go back to our basic LC circuit and modify it by, say, replacing the inductor by two inductors (of different inductivities) connected to the capacitor in parallel (the drawing uses this Mathematica code):
Now let us employ a tiny demon who will be periodically switching between the inductors back and forth, so that exactly one of the two inductors is included into the circuit at each time moment. If the values of the switching times, the two inductivities, and the capacity are in appropriate agreement between them (say as in the first of the above two counterexamples), then a self-resonance will occur, (almost) without any external force and (almost) without expending energy, with the applitudes increasing (almost) exponentially (almost) to infinity! Does this not contradict the energy conservation law?
