Forcing the uniqueness of a solution of an ODE For $n\geq 1$, $f_n\in\mathcal{C}^1([0,1],\mathbb{R})$ such that $f_n(x)\geq\sqrt{x}$ for $x\in[0,1]$, and
$$\lim\limits_{n\to+\infty}\sup_{x\in[0,1]}\big|f_n(x)-\sqrt{x}\big|= 0.$$
Let $y_n$ be the unique solution of
$$\begin{cases}
y_n(0)=0 \\
y_n'=f_n(y_n) \text{ on [0,1]}.
\end{cases}$$
Question: Is there a function $y\in\mathcal{C}^1([0,1],\mathbb{R})$ such that
$$\lim\limits_{n\to+\infty}\sup_{x\in[0,1]}\big|y_n(x)-y(x)\big|= 0$$
which is solution of the system (which has itself an infinity of solutions)
$$\begin{cases}
y(0)=0 \\
y'=\sqrt{y} \text{ on [0,1]}
\end{cases}$$
and satisfies the condition: $y(x)>0$ for $x\in\,]0,1]$ (which, I hope, permits to characterize $y$).
 A: $\newcommand\ep\varepsilon$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 inequalities $\ln(t+\ep)\le t+\ep-1<t$, 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.
A: 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 $\sqrt{\cdot}$ 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.
