Theorem. Let $U$, $V$, $W$ be normed linear spaces. Let $\Omega$ be open in $U$ and $\Upsilon$ be open in $V$. Let $f\colon \Omega\to\Upsilon$ be (Fréchet) differentiable at $c\in\Omega$ and $g\colon \Upsilon\to W$ be differentiable at $f(c)\in\Upsilon$. Then $g\circ f$ is differentiable at $c$ with $$ D(g\circ f)(c) = Dg(f(c))\circ Df(c)\text. $$
My proof:
Denote $L := Df(c)$ and $M := Dg(f(c))$. Let $\delta_1 > 0$ be such that for all $y\in B_{\delta_1}(f(c))\cap\Upsilon\setminus\{f(c)\}$, we have $$ \lVert g(y) - g(f(c)) - M(y - f(c)) \rVert < \epsilon \lVert y - f(c) \rVert.\tag{1} $$
Since $f$ will be continuous at $c$, take $\delta_2 > 0$ such that for all $x\in B_{\delta_2}(c)\cap \Omega\setminus\{c\}$, we have $$ \lVert f(x) - f(c) \rVert < \delta_2 $$ which due to (1) implies that $$ \lVert g(f(x)) - g(f(c)) - M(f(x) - f(c)) \rVert < \epsilon \lVert f(x) - f(c) \rVert.\tag{2} $$
W.l.o.g., assume that for all $x\in B_{\delta_2}(c)\cap\Omega\setminus\{c\}$, we have $$ \lVert f(x) - f(c) - L(x - c) \rVert < \epsilon\lVert x - c\rVert.\tag{3} $$
Now, for a general $x\in B_{\delta_2}(c)\cap\Omega\setminus\{c\}$, we have $$ \begin{align*} &\lVert g(f(x)) - g(f(c)) - ML(x - c)\rVert\\[1ex] & \le \lVert g(f(x)) - g(f(c)) - ML(f(x) - f(c))\rVert \\ & \phantom{abcdefghijklmn} + \lVert M(f(x) - f(c) - L(x - c))\rVert\\[1ex] & \le \epsilon\lVert f(x) - f(c) \rVert + \lVert M\rVert\, \lVert (f(x) - f(c) - L(x - c))\rVert & \text{by (2)}\\[1ex] & \le (\epsilon + \lVert M\rVert) \lVert f(x) - f(c) - L(x - c)\rVert + \epsilon\lVert L(x - c)\\[1ex] & < (\epsilon + \lVert M\rVert)\epsilon\lVert x - c\rVert +\epsilon\lVert L\rVert\, \lVert x - c\rVert & \text{by (3)}\\[1ex] &= \epsilon(\epsilon + \lVert M\rVert +\lVert L\rVert)\lVert x - c\rVert. \end{align*} $$
And we are virtually done.
Question: Do you spot any loose ends in the above proof?
By $ML$, I mean $M\circ L$.
Note that the spaces can be over $\mathbb R$ or $\mathbb C$ (but it's common for all, $U$, $V$, $W$), and possibly infinite dimensional.