Full question:
Suppose $f:U\to \mathbb{R}^m$ is differentiable and $U$ is convex. If there exists an $M>0$ such that $\vert Df(x)\vert\leq M$ $\forall x\in U$ Show that $\vert f(b)-f(a)\vert \leq M\vert b-a \vert$.
My proof:
Fix some arbitrary $a,b\in U$ Since $f:\mathbb{R}^n\to\mathbb{R}^m$, each component function $f_i$ is a function from $\mathbb{R}^n\to\mathbb{R}$ Define the function $\gamma:\mathbb{R}\to U$ by $\gamma (t)=\vec{a}+t(\vec{b}-\vec{a})$
Let $h(t):=f_i\circ\gamma (t)$, where $f_i$ is any component function of $f$. We can do this because U is convex and contains the straight line from a to b. Since $f$ is differentiable we know each $f_i$ is differentiable. And $h(t):\mathbb{R}\to\mathbb{R}$
Apply FTC for $h(1)-h(0)$ we get: $h(1)-h(0)=f_i(b)-f_i(a)=\int_0^1\frac{d}{dx}f_i(\gamma(t))dt=\int_0^1\nabla f_i(\vec{a}+t(\vec{b}-\vec{a}))\cdot (\vec{b}-\vec{a})dt$
then $\vert f_i(\vec{b})-f_i(\vec{a})\vert=\vert\int_0^1\nabla f_i(\vec{a}+t(\vec{b}-\vec{a}))\cdot(\vec{b}-\vec{a})dt\vert$
$\leq\int_0^1\vert\nabla f_i(\vec{a}+t(\vec{b}-\vec{a}))\cdot (\vec{b}-\vec{a})\vert dt$ $\leq \int_0^1 \vert\nabla f_i(\vec{a}+t(\vec{b}-\vec{a}))\vert\vert(b-a)\vert dt$
By assumption $\vert\nabla f_i(x)\vert\leq M$ thus $\vert f_i(\vec{b})-f_i(\vec{a})\vert\leq M\vert(\vec{b}-\vec{a})\vert\int_0^1dt=M\vert(\vec{b}-\vec{a})\vert$
then for $f:\mathbb{R}^n\to\mathbb{R}^m$ $\vert\vert f(\vec{b})-f(\vec{a})\vert\vert=\sqrt{\sum_{i=1}^m(f_i(\vec{b})-f_i(\vec{a}))^2}\leq\sqrt{M^2\vert(\vec{b}-\vec{a})\vert^2(m)}=\sqrt{m}M\vert(\vec{b}-\vec{a})\vert$
The problem here is that I get this $\sqrt{m}$, the solution apparently should be that you can show its bounded just by $M\vert b-a \vert$.
Consider the case of a vector $v$ where $\|v\| \leq 1$. Then each $|v_i| \leq 1$ and so $\|v\| = \sqrt{v_1^2+\cdots+v_n^2} \leq \sqrt{n\cdot 1} = \sqrt{n}$. True, but we've lost some precision! That's where your square root came in and you can find exactly in your proof where that happened. So a different approach, using the total derivative will be necessary.
Adapted from Sternberg's Advanced Calculus you can argue as follows. Call $h(t) = f(\gamma(t))$ so $h:[0,1]\rightarrow \mathbb{R}^m$ and by the chain rule $\|h'(t)\| \leq M|b-a|$. Fix $\epsilon > 0$. Consider the "good set" $$G = \{t\in[0,1]:\; |h(t)-h(0)| \leq (M|b-a|+\epsilon)t + \epsilon \}$$ If we show that $1\in G$ then since $\epsilon$ is arbitrary, the result follows (You'll see why that multiplicative $\epsilon$ appears in a minute).
Firstly note by continuity of the map $t\mapsto \|h(t)-h(0)\|-(M|b-a|+\epsilon)t$ at zero, the set $G$ must contain some interval to the right of $0$. Let $t^* = \sup G$ and again by continuity it is seen that $t^* \in G$. If (for contradiction) $t^* \neq 1$ then by definition of the derivative $h'(t^*)$:
$$ \frac{\|h(t)-h(t^*)\|}{t-t^*} \leq M|b-a| + \epsilon $$ for all $t\in (t^*,t^*+\delta)$ for some $\delta>0$. Then for such $t$: \begin{align*} \|h(t)-h(0)\| &\leq \|h(t)-h(t^*)\|+\|h(t^*)-h(0)\| \\ &\leq (M|b-a|+\epsilon)(t-t^*) + (M|b-a|+\epsilon)t^* +\epsilon\\ &=(M|b-a|+\epsilon)t+\epsilon \end{align*} But that means $t$ belongs to $G$, a contradiction since $t>t^*$.