If $G\in C([0,1])$ and strictly increasing, can we find a sequence $G_n\n C^{\infty}$ with uniformly equicontinuous density?

Let $G:[0,1]\rightarrow [0,1]$ be a strictly increasing and continuous cdf with $G(1)=1$.

I have proven some property for $G\in C^{\infty}([0,1])$ that relies on the continuity of $g(x)=G'(x)$. I hope to generalize the result for $G\in C([0,1])$.

Since $C^{\infty}([0,1])$ is dense in $C([0,1])$, we can find a sequence $\{G_n\}\subset C^{\infty}([0,1])$ converging to $G\in C([0,1])$.Denote $g_n=G_n'(x)$. Morover, since $G$ is strictly increasing, it is almost everywhere differentiable although $g'(x)$ need not be continuous.

I came across a Conditions for convergence of a derivative, given the function itself is convergent that shows that if $\{g_n\}$ is uniformly equicontinuous, then $g_n\rightarrow g$ uniformly.

My attempt was the following which clearly doesn't work:

Since $G_n\in C^{\infty}([0,1])$, each $g_n$ is Lipschitz (though the constant may differ for $n$). However, this doesn't imply $\{g_n\}$ is uniformly Lipschitz.

I have the following few questions.

1. Can we find $\{G_n\}$ so that $\{g_n\}$ is uniformly equicontinuous?
2. Can we find $\{G_n\}$ so that $\{g_n\}$ is an monotonically increasing/decreasing sequence in $n$ for all $x$? i.e. $g_n(x)\geq g_{n+1}(x)$ for all $n\in \mathbb{N}, x\in[0,1]$ (or $g_n(x)\leq g_{n+1}(x)$).
3. If answers to either (or both) question is "no", what additional sufficient conditions do we need to insure the existence of such sequences?

Any comments would be helpful. Thank you.