Proof of the derivative of the compostion of functions. This question comes from the proof of the derivative of a composite function which I lifted straight from Proof Wiki 
https://proofwiki.org/wiki/Derivative_of_Composite_Function:
Let $f, g, h$ be continuous real functions such that: $\forall x \in \mathbb R: h \left({x}\right) = f \circ g \left({x}\right) = f \left({g \left({x}\right)}\right)$ 
Then: $h' \left({x}\right) = f' \left({g \left({x}\right)}\right) g' \left({x}\right)$ 
where $h'$ denotes the derivative of $h$.
Proof:
Let $g \left({x}\right) = y$, and let
$g(x+δx)  = y+δy$
Thus: 
Thus:
$δy→0\ \ \ \ \ $   as    $\  \ \ \ \ δx→0$
and:
$\frac{δy}{δx}→g′(x) \ \ \ \ \ \ \ \ \ (1)$
Case 1:
Suppose $g′(x)≠0$ and that $δx$ is small but non-zero.
Then $δy≠0$ from (1) above, and:
$lim_{δx→0} \frac{h(x+δx)−h(x)}{δx} =    lim_{δx→0}\frac{f(g(x+δx))−f(g(x))}{g(x+δx)−g(x)}$ $\frac{g(x+δx)−g(x)}{δx}$
=               $lim_{δx→0}\frac{f(y+δy)−f(y)}{δy}$ $\frac{δy}{δx}$
=               $f′(y)g′(x)$
My question: Why does the assumption that $g'(x)$ $\neq$ $0$ imply that $dy$ $\neq$ $0$ and why does this in turn imply that the expression $\ \ \ \ \ lim_{δx→0} \frac{h(x+δx)−h(x)}{δx}\ \ \ \ \ $      
is equal to $\ \ \ \ \ \ \ \ \ \ \ \ \ \ lim_{δx→0}\frac{f(g(x+δx))−f(g(x))}{g(x+δx)−g(x)}\frac{g(x+δx)−g(x)}{δx}$      

If $g'(x)$ $\neq$ $0$ 
Since $g'(x) = \frac{dy}{dx}$, and we have also assumed $dx$ is small but non-zero. So, the product of two non-zero values yield non-zero result.  implies that $dy$ $\neq$ $0$
Since $dy$ $\neq$ $0$, by our definition of $dy$, we know this is equivalent to saying  ${g(x+δx)−g(x)}$$\neq$ $0$. Then, this means it's valid to multiply top and bottom of the fraction by ${g(x+δx)−g(x)}$. Which in turn yields:
The expression$\ \ \ \ \ lim_{δx→0} \frac{h(x+δx)−h(x)}{δx}\ \ \ \ \ $      
is equal to $\ \ \ \ \ \ \ \ \ \ \ \ \ \ lim_{δx→0}\frac{f(g(x+δx))−f(g(x))}{g(x+δx)−g(x)}\frac{g(x+δx)−g(x)}{δx}$      
