Is this proof of the chain rule valid ?
We want to find the value of : $$(g\circ f)'(x) =\lim\limits_{h \to ~0}\frac{(g\circ f)(x + h) - (g\circ f)(h)}{h} \tag{1}$$
But we know from the mean value theorem that there is an $a$ in $[f(x),f(x+h)]$ such that $$g(f(x+h))-g(f(x))=(f(x+h))-f(x))g'(a)$$ so plugging in (1) we obtain : $$(g\circ f)'(x) =\lim\limits_{h \to ~0}g'(a)\frac{f(x + h) - f(h)}{h} =g'(f(x))f'(x)$$
since when h tends to zero, $[f(x),f(x+h)]$ shrinks to $f(x)$.
Edit : from the comments, it appears that the proof is valid but leads to a weaker theorem than the usual chain-rule theorem, since the proof assumes the derivative of g to be continuous.