Is the following proof of the chain rule correct?
$\textbf{Chain rule}$: Consider the functions $f:A \subseteq \mathbb{R} \rightarrow B \subseteq \mathbb{R}$ and $g:B \subseteq \mathbb{R} \rightarrow \mathbb{R}$. Suppose that $a \in A$ is an accumulation point of $A$ and that $f$ is differentiable in $a$. Also suppose that $f(a)$ is an accumulation point of $B$ and that $g$ is differentiable in $f(a)$. Then $g \circ f$ is differentiable in $a$ and $(g \circ f)'(a)=g'(f(a))f'(a)$.
$\textbf{Proof}$: For every $x \in A \backslash \{ a \}$ is $\frac{(g \circ f)(x)-(g \circ f)(a)}{x-a} = \frac{g(f(x))-f(f(a))}{x-a} =\frac{g(f(x))-g(f(a))}{f(x)-f(a)} \frac{f(x)-f(a)}{x-a}$. We know that $\lim_{x \rightarrow a} \frac{f(x)-f(a)}{x-a} =f'(a)$. Because of continuity of $f$ in a, $f(x) \rightarrow f(x)$ if $x \rightarrow a$ and thus will $\lim_{x \rightarrow a} \frac{g(f(x))-g(f(a))}{f(x)-f(a)} = \lim_{y \rightarrow a} \frac{g(y)-g(f(a))}{y-f(a)} = g'(f(a))$. Because the limit of the product of two functions is the product of the limit of the functions, the formula has been proven.