The sharp general result in this direction is the classical law of the iterated logarithm (LIL). Suppose, after renormalizing if necessary, that the $x_n$ are iid with zero mean and unit variance. Then the LIL states that $$\limsup_{n \to \infty} \frac{X_N}{\sqrt{N \log \log N}} = \sqrt{2}, \quad \text{a.s.}$$ In your language, that says that with probability 1, $X_N$ is $O(\sqrt{N \log \log N})$, and that this cannot be improved to $O(\sqrt{N})$.

To be more careful, it says that for $P$-almost every $\omega$, there is a finite number $C(\omega)$ such that $|X_N(\omega)| \le C(\omega) \sqrt{N \log \log N}$ for all $N$.

The sharp general result in this direction is the classical law of the iterated logarithm (LIL). Suppose, after renormalizing if necessary, that the $x_n$ are iid with zero mean and unit variance. Then the LIL states that $$\limsup_{n \to \infty} \frac{X_N}{\sqrt{N \log \log N}} = \sqrt{2}, \quad \text{a.s.}$$ In your language, that says that with probability 1, $X_N$ is $O(\sqrt{N \log \log N})$, and that this cannot be improved to $O(\sqrt{N})$.

To be more careful, it says that for $P$-almost every $\omega$, there is a finite number $C(\omega)$ such that $|X_N(\omega)| \le C(\omega) \sqrt{N \log \log N}$ for all $N$.

LIL isn't a direct corollary of CLT, and I believe there are settings where either may hold while the other fails. So I don't think it's true that the CLT "implies" the result you desire, but in any case they are both true in your setting.