Existence of a pointwise convergent subsequence

No, there does not. Consider $f_n(x) = \cos(4^n \pi x)$ on $[0,1]$ (with range in $[-1,1]$ not $[0,1]$, but you can transform it). Then for any subsequence $f_{n_k}$, there is some $x \in [0,1]$ such that $f_{n_k}(x) \ge \cos(\pi/4)$ if $k$ is odd and $\le -\cos(\pi/4)$ if $k$ is even. All you need to do is choose the base-$4$ digits of $x$ correctly.

EDIT: Any choice of the first $n$ digits after the "decimal" point leaves $x$ in an interval of length $4^{-n}$ on which $f_n$ goes either from $1$ to $-1$ or $-1$ to $1$. Choose the $n+1$'th digit to be $0$ or $3$ to make $f_n(x) \ge \cos(\pi/4)$ or $\le -\cos(\pi/4)$ (which is which depends on the previous choices of digits). For example, if you choose the first $3$ digits to be $0,3,0$, that says $x$ is between $0 \times 4^{-1} + 3 \times 4^{-2} + 0 \times 4^{-3} = 3/16$ and $0 \times 4^{-1} + 3 \times 4^{-2} + 1 \times 4^{-3} = 13/64$, say $x = 3/16 + 4^{-3} t$, $0 \le t \le 1$. On this interval, $f_3(x) = \cos(4^3 \pi x) = \cos(\pi t)$ goes from $1$ at $t=0$ to $-1$ at $t=1$. If you want $f_3(x) \ge \cos(\pi/4)$, choose the $4$'th digit to be $0$, if you want $f_3(x) \le - \cos(\pi/4)$ choose the $4$'th digit to be $3$.

A similar example using more machinery hence perhaps less tricky in the details: Let $X=[0,2\pi]$, $f_n(t)=\cos(nt)$. If $f_{n_k}(x)\to f(x)$ for (almost) every $x$ then Dominated Convergence shows that $||f_{n_k}-f||_2\to0$; this is impossible by orthogonality (for example, $||f_n-f_m||_2^2=2\pi$.)