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 $y'=\sqrt{y}$ on $[0,1]$ which vanishes only at $0$ is $y_0 : x \mapsto $(x/2)^2$.
The idea is to prove that the sequence $(y_n)_{n \ge 1}$ is equicontinuous and bounded below by $y_0$. Since they vanish at $0$, this sequence is relatively compact in $C([0,1],R)$. Then we see that the only limit point is $y_0$.
To be detailed later.
Next, for $n \ge 1$, and $x \in [0,1]$