Applying Arzelà–Ascoli Theorem 
Let $(g_n)$ be a sequence of twice differentiable functions defined on $[0,1]$, and assume that $g_n(0)=g_n'(0)$ for all $n$. Suppose also that $|g_n'(x)|\leq 1$ for all $n\in\mathbb{N}$ and all $x\in[0,1]$. Prove that there is a subsequence of $(g_n)$ converging uniformly on $[0,1]$.

This problem is screaming at me Arzelà-Ascoli Theorem. However, I am neither sure how to show that $(g_n)$ is uniformly bounded nor how to show that $(g_n)$ is equicontinuous.
Any help would be appreciated.

Edit: $(g_n)$ is uniformly bounded since $$|g_n(x)|=\left|g(0)+\int_0^xg_n'(x)dx\right| \leq |g_n(0)|+\int_0^x|g_n'(x)|dx \leq |g_n(0)|+ \int_0^x dx=1+x\leq2.$$
 A: For equicontinuity, you have that for some $\zeta \in [0,1]$,
$$
|g_n(x) - g_n(y)| = |g'_n(\zeta)||x-y| \le |x-y|
$$
by Taylor's theorem, so that $\{ g_n \}$ are all Lipschitz with constant $1$, hence equicontinuous. I don't know why you need twice differentiability ; $C^1$ is enough here.
For uniformly bounded, 
$$
|g_n(x)| = |g_n(x) - g_n(0) + g_n(0)| = |g_n'(\zeta)x + g_n'(0)| \le |g'_n(\zeta)||x|+|g'_n(0)| \le 1+x.
$$
again by Taylor's theorem and since $|g'_n(x)| \le 1$, plus with $x \ge 0$ I can remove the $| \cdot |$.
Hope that helps,
A: Uniformly bounded: For any $n$, we have $$|g_n(x)|=\left|g_n(0)+\int_0^x g_n'(t)dt\right|\leq |g_n(0)|+\int_0^x|g_n'(t)|dt\leq1+\int_0^xdt = 1+x$$
establishing that the sequence is uniformly bounded.
Equicontinuous: Let $\epsilon>0$ and set $\delta=\epsilon$. If $|x-y|<\delta$, then $$|g_n(x)-g_n(y)|=\left|\int_x^yg_n'(t)dt\right|\leq \left|\int_x^y|g'(t)|dt\right|\leq \left|\int_x^ydt\right|=|x-y|<\delta=\epsilon$$
thus the sequence $(g_n)$ is equicontinuous.
A: The fact that the sequence $(g_n)$ is equicontinuous follows immediately from the mean value theorem, and the assumption $|g_n'(x)|\leq 1$. Since $|g_n(x)-g_n(y)|=|g_n'(\xi)|\cdot|x-y|\leq |x-y|$ (where $\xi$ is some element in $(x,y)$). 
