- For boundedness: Fix $n$; there is $x_n$ such that $f_n(x_n)=0$ by the first condition. Then $$|f_n(x)|\leqslant\left|\int_x^{x_n}|f_n'(t)|dt\right|\leqslant\int_0^1t^{-1/2}dt.$$
- For equi-continuity, the second condition gives that $|f_n(x)-f_n(y)|\leqslant 2\sqrt{|x-y|}$.
Note that since $f'_n$ is not assumed continuous, we can't apply directly fundamental theorem of analysis. But we can use the fact that $f_n$ is absolutely continuous, with a good condition on the derivative, to get what we want.
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
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: $$|f_n(x)|=\left|f(0)+\int_0^xf_n'(t)dt\right|\leq 0+\int_0^x|f_n(t)|dt\leq \int_0^xMdt=Mx\leq M$$