This question would be possibly at a better place on MathStack Exchange. 

Yet, once the statement of the question is corrected (the functions $y_n$ need to be defined on $\mathbb{R_+}$ and not only on $[0,1]$, and they should converge uniformly to $f$ on compact sets), the answer will be positive. 

For strictly positive functions $y'=\sqrt{y}$ is equivalent to $y'/\sqrt{y}=1$, namely $2\sqrt{y}-x$ constant (by integration). Hence the only positive solution of the ODE $y'=\sqrt{y}$ on $[0,1]$ which vanishes only at $0$ is $y_0 : x \mapsto (x/2)^2$. 

In a same way, the only positive solution of the ODE $y'=1+\sqrt{y}$ on $\mathbb{R}_+$ which vanishes at $0$ is $G^{-1}$, where $G$ is the strictly (increasing) function defined by 
$$G(y) := \int_0^y \frac{\mathrm{d}z}{1+\sqrt{z}}.$$ 

Since the $(f_n)_{n \ge 1}$ are $\mathcal{C}^1$ and bounded below by $\sqrt{}$, they are strictly positive everywhere (including at $0$), so the solutions $y_n$ of the Cauchy problem $y'=f_n(y)$, $y(0)=0$ are (strictly) increasing. 

Let $A:=G^{-1}(1)$. We assume that $(f_n)_{n \ge 1}$ converges uniformly to $\sqrt{}$ on $[0,A]$. Hence, whenever $n$ is greater to some positive integer $N$, we have $\sqrt{} \le f_n \le 1+\sqrt{}$ on $[0,A]$. While $y_n$ remains in $]0,A]$, we derive 
$$\frac{y_n'}{1+\sqrt{y_n}} \le 1 \le \frac{y_n'}{\sqrt{y_n}}.$$ 
Since $y_n(0)=0$, we get by integrating
$$G(y_n(x)) \le x \le 2\sqrt{y_n(x)},$$
so $y_0(x) = (x/2)^2 \le y_n(x) \le G^{-1}(x)$. 
As a result, for every $n \ge N$, $y_n$ remains on $[0,A]$ on the time interval $[0,1]$ and $y_n'$ is uniformly bounded (namely $0 \le y_n' \le 1+\sqrt{A}$ on the time interval $[0,1]$. By Ascoli-Arzela's theorem, the sequence $(y_n)_{n \ge N}$ is relatively compact in $\mathcal{C}([0,1],\mathbb{R})$. 

Thus, we have to check that its only limit point is $y_0$. Indeed, if $y$ is a subsequential limit, then taking limits along this subsequence in the equalities 
$$y_n(x) = \int_0^x f_n(y_n(t)) \mathrm{d}t$$
yields 
$$y_n(x) = \int_0^x \sqrt{y_n(t)} \mathrm{d}t$$
But $y_n$ is bounded below by $y_0$. Hence $y_n=y_0$ by the uniqueness argument viewed at the beginning. We are done.