Prove that in these conditions, $f(x)<g(x):\forall x\in[0,1)$

Let $f,g:[0,1]\longrightarrow\mathbb{R}$ be two continuous and differentiable functions in $[0,1]$ such that:

$$f(0)=0$$ $$g(0)=2$$ $$|f'(x)|,|g'(x)|\leq1:\forall x,y\in[0,1]$$

Then prove that $f(x)<g(x):\forall x\in[0,1)$ and $f(1)\leq g(1)$

I can see the geometric interpretation of this to be pretty straightforward, since we know that $g'(x)\geq-1$ and $f'(x)\leq1$, and they are two units apart at the start point, even considering the most extreme case they would only intersect in $x=1$.

However I'm trying to find an analytical proof for this. I have considered $$h(x):=g(x)-f(x)$$

And trying to prove $h(x)\geq0:\forall x\in[0,1)$, but I can't quite get it. I have also thought that maybe Lagrange's theorem could be useful as the function is continuous and differentiable, but I'm not sure how to apply it to $h(x)$.

Note that $h(0) = 2 > 0$. Suppose that $h(c) = 0$ for some $c \in [0, 1)$. By the mean value theorem, we could then find a point $\alpha$ between $0$ and $c$ for which
$$h'(\alpha) = \frac{h(0) - h(c)}{0 - c} = \frac{2}{c} > 2$$
since $c < 1$. Do you see how this leads to a contradiction? Can you finish the proof from here?