In here, since I want to use Arzela Ascoli, but I am not sure how to prove boundedness for this sequence. For equicontinuity, since fn is continuous then, I could apply this answer however I do not understand the proof very clearly. Could someone please explain it to me?
Let's show uniform boundedness.
Take $x\in [0,1]$, choose a decreasing sequence $a_j\rightarrow 0^+$ for $j\rightarrow +\infty$ and $a_1=x$. Fix $j>0$ such that $a_j<x$. Now fix $\epsilon>0$ and a partition $\{ a_j=t_0<t_1<....<t_k=x\}$ such that $\sum_{m=0}^{k-1}\frac{1}{\sqrt{t_m}}(t_{m+1}-t_m)\leq\int_{a_j}^{x}\frac{1}{\sqrt{s}}ds+\epsilon$ (it's just the definition of Riemann integral). Now using Lagrange theorem we compute:
$|f_n(x)|\leq \sum_{m=0}^{k-1} |f_n(t_{m+1})-f_n(t_m)|+|f_n(a_j)|=\sum_{m=0}^{k-1} |f_n'(y_m)|(t_{m+1}-t_{m})+|f_n(a_j)|\leq \sum_{m=0}^{k-1}\frac{1}{\sqrt{y_m}}(t_{m+1}-t_{m})+|f_n(a_j)|\leq\sum_{m=0}^{k-1}\frac{1}{\sqrt{t_m}}(t_{m+1}-t_{m})+|f_n(a_j)|\leq \int_{a_j}^{x}\frac{1}{\sqrt{s}}ds+|f_n(a_j)|+\epsilon=2(\sqrt{x}-\sqrt{a_j})+|f_n(a_j)|+\epsilon\leq 2(1-\sqrt{a_j})+|f_n(a_j)|+\epsilon$
because $y_m\in(t_m,t_{m+1})$. Now for $j\rightarrow +\infty$, because of continuity of $f_n$ and $f_n(0)=0$ we get:
$|f_n(x)|\leq 2+\epsilon$
and being $\epsilon$ arbitrary $|f_n(x)|\leq 2$. Because $x\in[0,1]$ was arbitrary we get $|f_n|\leq 2$ for all $n$.
Now to show equicontinuity i think we can use a similar argument.