Intuitive Proof of the Chain Rule in 1 Variable
If $\frac{du}{dx}=k\neq 0$ at some $x$, then a first-order (that is: linear) approximation of $du$ close to $x$, gives $$ du=k\cdot dx \Rightarrow \frac{1}{du}=\frac{1}{k\cdot dx} $$ thus: $$ \frac{dy}{du}\cdot\frac{du}{dx}=\frac{dy}{k\cdot dx}\cdot k=\frac{dy}{dx} $$ Intuitively, you should be thinking of differentials as "small changes". So small, that even linear approximation would be good "enough".
If you want intuitive and simple:
$$\frac{dy}{dx}=\frac{dy}{\color{#4499de}{du}}\frac{\color{#4499de}{du}}{dx}$$
where the $du$'s cancel out.
If you want to be more rigorous, replace $dy,dx,du$ with $\Delta y,\Delta x,\Delta u$, which is the change with respect to $x$, and take the limit as $\Delta\to0$, which becomes the derivatives.
Some intuition: If $f(x) = m_1x + b_1, g(x) =m_2x + b_2,$ then $(g\circ f)(x) = m_2m_1x + (m_2b_1 + b_2).$ So in the case of linear functions, the slope of their composition is the product of the slopes. Now if $f,g$ are differentiable at $a, f(a)$ respectively, we can expect that, near $a,$ $g\circ f$ is very close to the composition of their tangent lines. Thus the slope of their composition at $a$ should be the product of the two slopes, i.e., $(g\circ f)'(a) = g'(f(a))\cdot f'(a).$