$\newcommand\ep\varepsilon$First, the conditions that $f_n\in\mathcal{C}^1([0,1],\mathbb{R})$ and $f_n(x)\ge\sqrt{x}$ for $x\in[0,1]$ imply $f_n(0)>0$. Since 
\begin{equation*}
	\begin{cases}
y_n(0)=0, \\
y_n'=f_n(y_n) \text{ on [0,1]}, 
\end{cases}
\tag{2}\label{2}
\end{equation*}
we see that $y_n>0$ in a right neighborhood of $0$. Since $y'_n=f_n(y_n)\ge0$, we see that $y_n>0$  on $(0,1]$. 
Letting then $u_n:=\sqrt{y_n}$, we get $2u_n u'_n=f_n(u_n^2)\ge u_n$, whence $u'_n\ge1/2$, $u_n(x)\ge x/2$, and 
\begin{equation*}
	y_n(x)\ge x^2/4 \tag{3}\label{3}
\end{equation*}
for all $x\in[0,1]$. 

Next, for $\ep\in(0,1)$, let $z_\ep$ be the unique solution of the ODE 
\begin{equation*}
	z'_\ep=\sqrt{z_\ep}+\ep
\end{equation*}
on $[0,1]$ with the initial condition $z_\ep(0)=0$. It is not hard to see that 
\begin{equation*}
	z_\ep(x)\to x^2/4 \tag{4}\label{4}
\end{equation*}
uniformly in $x\in[0,1]$ as $\ep\to0$. (See details on this at th end of this answer.)

Let now 
\begin{equation*}
	\ep_n:=\sup_{x\in[0,1]}\big|f_n(x)-\sqrt{x}\big|
	=\sup_{x\in[0,1]}\big(f_n(x)-\sqrt{x}\big), 
\end{equation*}
so that $\ep_n\to0$ (as $n\to\infty$), and then let 
\begin{equation*}
	w_n:=z_{\ep_n+1/n}. 
\end{equation*}
So, $y_n(0)=0=w_n(0)$, 
\begin{equation*}
	y_n'\le\sqrt{y_n}+\ep_n,\quad w_n'=\sqrt{w_n}+\ep_n+1/n. 
\end{equation*}
Suppose that 
\begin{equation*}
	x_n:=\sup\{x\in[0,1]\colon y_n\le w_n\text{ on }[0,x]\}<1. 
\end{equation*}
Then $x_n>0$ and $w_n(x_n)=y_n(x_n)$, and hence $y'_n(x_n)\ge w'_n(x_n)$, so that 
\begin{equation*}
	\sqrt{w_n(x_n)}+\ep_n=\sqrt{y_n(x_n)}+\ep_n \\ 
	\ge y'_n(x_n)\ge w'_n(x_n)=\sqrt{w_n(x_n)}+\ep_n+1/n,
\end{equation*}
a contradiction. So, $x_n=1$ and hence, in view of \eqref{3}, 
\begin{equation*}
	x^2/4\le y_n(x)\le w_n(x)=z_{\ep_n+1/n}(x)\to x^2/4
\end{equation*}
uniformly in $x\in[0,1]$, by \eqref{4}. 

On the other hand, the only solution $y$ of the system  
\begin{equation*}
	\begin{cases}
y(0)=0 \\
y'=\sqrt{y} \text{ on [0,1]}
\end{cases}
\end{equation*}
such that $y>0$ on $(0,1]$ is given by the formula $y(x)=x^2/4$. 

Thus, $y_n\to y$ uniformly on $[0,1]$, as desired.  

---

**Details on \eqref{4}:** Letting $t_\ep:=\sqrt{z_\ep}$, rewrite 
$z'_\ep=\sqrt{z_\ep}+\ep$ as $2t_\ep t'_\ep=t_\ep+\ep$. "Separating the variables", we find that $t_\ep=g^{-1}_\ep$, where 
\begin{equation*}
	g_\ep(t):=2t-2\ep\ln\frac{t+\ep}\ep. 
\end{equation*}
We have $g'_\ep(t):=2-\frac{2\ep}{t+\ep}0$ for $t>0$, so that the inverse function $g^{-1}_\ep$ is well defined. Using the inequality $\ln(t+\ep)\le t+\ep-1$, we get 
\begin{equation*}
	g_\ep(t)\ge (2-2\ep)t+2\ep\ln\ep  
\end{equation*}
for $t\ge0$, whence for $\ep\to0$ we have $t_\ep(x)=g^{-1}_\ep(x)\le\frac{x-2\ep\ln\ep}{2-2\ep}\to x/2$ and 
\begin{equation*}
	z_\ep(x)=t_\ep(x)^2\le(1+o(1))x^2/4
\end{equation*}
uniformly in $x\in[0,1]$. Also, similarly to \eqref{3}, $z_\ep(x)\ge x^2/4$ for $x\in[0,1]$. Now \eqref{4} follows.