Rudin's definition of differentiability states the following:
Suppose $E$ is an open subset of $\mathbb{R}^n$. Let $f: E \to \mathbb{R}^m$ be a map. Let $x \in E$. We say that $f$ is differentiable at $x$ if there is a linear transformation $A: \mathbb{R}^n \to \mathbb{R}^m$ such that
$$\lim_{h \to 0} \frac{\Vert f(x+h)-f(x)-Ah\Vert}{\Vert h \Vert} = 0$$
Later, the multivariable chain rule is proven:
Suppose $E$ is an open subset of $\mathbb{R}^n$, $f$ is differentiable at $x_0$, $g$ maps an open set containing $f(E)$ into $\mathbb{R}^k$, and $g$ is differentiable at $f(x_0)$. Then the mapping $F$ of $E$ into $\mathbb{R}^k$ defined by $F(x) = g(f(x))$ is differentiable at $x_0$ and $F(x_0) = g'(f(x_0))f'(x_0)$.
Theorem 9.19 states:
Suppose $f$ maps an open convex set $E \subseteq \mathbb{R^n}$ into $\mathbb{R}^m$, $f$ is differentiable on $E$, and there is a real number $M \geq 0$ such that $\Vert f'(x) \Vert \leq M$ for all $x \in E$. Then
$$\Vert f(b)- f(a) \Vert \leq M \Vert b-a \Vert$$ for all $a, b \in \mathbb{R}^n$
This is the beginning of the proof:
Fix $a, b\in E$. Define $\gamma(t) = (1-t)a + tb$ for all $t \in > \mathbb{R}$ such that $\gamma(t) \in E$. Since $E$ is convex, $\gamma(t) \in E$ if $0 \leq t \leq 1$.
Put $g(t):= f(\gamma(t))$.
Then $g'(t) = f'(\gamma(t))\gamma'(t) = f'(\gamma(t))(b-a)$ so that $\Vert g'(t) \Vert \leq \Vert f'(\gamma(t)) \Vert\Vert b-a \Vert \leq M \Vert b-a \Vert$ for all $t \in [0,1]$ [and the proof continues...]
Apparently, the chain rule is used. However, mustn't the domain of $\gamma$ be an open subset of $\mathbb{R}$ in order to apply the chain rule? In particular, I can see that the conclusion might hold for $t \in (0,1)$, but what about $t=0, t=1?$
Since $E$ is open, not only is $\gamma(t) \in E$ for $0 \le t \le 1$, there exists $\epsilon_0 > 0$ with the property that $\gamma(t) \in E$ for $-\epsilon \le t \le 1 + \epsilon$ whenever $0 < \epsilon < \epsilon_0$.
By your reasoning above, for any such $\epsilon$ you have $$\|\gamma'(t)\| \le M \|\gamma(1+\epsilon) - \gamma(-\epsilon)\|$$ if $-\epsilon < t < 1+\epsilon$, and in particular if $t \in [0,1]$. Since $$\gamma(1+\epsilon) - \gamma(-\epsilon) = [- \epsilon a + (1+\epsilon)b] - [(1+\epsilon) a -\epsilon b] = (b-a) + 2\epsilon(b-a)$$ it follows that $$\|\gamma'(t)\| \le M (\|b-a\| + 2\epsilon \|b-a\|)$$ for all $t \in [0,1]$. Now let $\epsilon \to 0^+$.