As stated, the sequence need not be uniformly (in your words, pointwise, but I have not heard this term used this way before) bounded, and in fact need not have a uniformly convergent subsequence. Take for example the sequence of functions $f_n(x)=n$. I suspect that you are missing an additional assumption, of the form "each $f_n$ satisfies $f_n(0)=0$". If this is the case, then we can apply the Fundamental Theorem of Calculus:
$$|f_n(x)|=\left|f(0)+\int_0^xf_n'(t)dt\right|\leq 0+\int_0^x|f_n(t)|dt\leq \int_0^xMdt=Mx\leq M$$
Note that any subsequence of $(f_n)_{n\in\mathbb N}$ converges pointwise the function $$f(x)\equiv\begin{cases}0&\text{if $x\in[0,1)$,}\\1&\text{if $x=1$.}\end{cases}$$ Hence, if uniform convergence occurred, then the uniform limit would have to agree with the pointwise limit. However, the pointwise limit $f$ is not continuous. Since the uniform limit of continuous functions is continuous, it follows that no subsequence can converge uniformly.
Now, $\|f_n\|=\sup_{x\in[0,1]}|x^n|=1$ for each $n\in\mathbb N$, so that the sequence $(f_n)_{n\in\mathbb N}$ is contained in the unit ball of $C([0,1])$ (when this space is endowed with the uniform norm). If the unit ball were compact, there would exist a uniformly convergent subsequence, which, as one has seen, is not possible.
The Arzelà–Ascoli theorem doesn't work here because the sequence is not equicontinuous at $x=1$. To see this, pick any $\varepsilon\in(0,1)$. If the sequence were equicontinuous, there would exist some $\delta>0$ such that if $y\in(\max\{1-\delta,0\},1]$, then $$|f_n(1)-f_n(y)|<\varepsilon\quad\forall n\in\mathbb N.$$ But this is impossible, as $$|f_n(1)-f_n(y)|=|1-y^n|\to 1\quad\text{as $n\to\infty$}$$ whenever $y\in(0,1)$, so $|f_n(1)-f_n(y)|$ eventually exceeds $\varepsilon<1$.
It may be worth keeping in mind that, in general, if $(X,\|\cdot\|)$ is any infinite-dimensional normed vector space, then its closed unit ball is never compact with respect to the norm topology.
Best Answer
@Nate Eldredge already said it all: \begin{align*} |f(z)-f(y)|&=\left|\int_{y}^{z}f'(x)dx\right|\\ &\leq\int_{y}^{z}|f'(x)|dx\\ &=\int_{0}^{1}|f'(x)|\chi_{[y,z]}dx\\ &\leq\left(\int_{0}^{1}|f'(x)| ^{2}dx\right)^{1/2}|z-y|^{1/2}, \end{align*} for $z\geq y$, the rest should be straight forward.