I'm trying to prove for twice-differentiable $f:[0,\infty) \to \mathbb{R}$ that if $f \in L^2([0,\infty))$ and $f'' \in L^2([0,\infty))$, then follows that $f' \in L^2([0,\infty))$.
I know by Cauchy-Schwartz that
$$\left|\int_0^\infty ff'' \right| \leq \int_0^\infty |ff''| \leq \sqrt{\int_0^\infty|f|^2} \sqrt{\int_0^\infty |f''|^2}< \infty$$
I also think I can use integration by parts (is this true for Lebesgue integrals?):
$$\int_0^xff'' = f(x)f'(x) - f(0)f'(0) - \int_0^x|f'|^2$$
Since $\lim_{x \to \infty}\int_0^x ff'' = \int_0^\infty ff''$ exists I can claim that $\lim_{x \to \infty}\int_0^x |f'|^2 < \infty$ if I know that $\lim_{x \to \infty} f(x) f'(x)$ exists.
I think that $\lim_{x \to \infty} f(x) f'(x) \neq \pm \infty$ but how can I be sure the limit exists at all? Can I still use this line of reasoning?
Your approach is fine.
To finish, note that $x \mapsto \int_0^x |f'|^2$ is nondecreasing and either $\lim_{x \to \infty} \int_0^x |f'|^2$ exists and is finite or the integral diverges to $+\infty$.
We have
$$f(x)f'(x) = f(0)f'(0) + \int_0^x |f'|^2 + \int_0^xff''$$
As shown, the second integral on the RHS converges. Consequently, $f(x)f'(x)$ either approaches a finite limit or tends to $+\infty$ as $x \to \infty$.
If the finite limit exists, then we are done.
On the other hand, if $f(x)f'(x)$ tends to $+\infty$ then $\int_0^\infty ff'$ is divergent and
$$\lim_{x \to \infty}|f(x)|^2 = |f(0)|^2+\lim_{x \to \infty}\int_0^x (f^2)' = |f(0)|^2+\lim_{x \to \infty}2 \int_0^x f f' = +\infty.$$
This implies that $\int_0^\infty |f|^2$ diverges to $+\infty$ -- contradicting the premise $f \in L^2([0,\infty)).$