Theorem Let $A \subseteq \mathbb R^n$ be open and $f : A\to \mathbb R^m$ differentiable on $A$. If we have $$\Vert Df\Vert_\infty := \sup\{\Vert Df(x)\Vert_o: \ x\in A\}<\infty$$ Then for any segment $[x,y] = \{(1-t)x+ty:t \in [0,1]\}\subseteq A$, $$\Vert f(y) - f(x)\Vert \le \Vert Df\Vert_\infty \Vert y-x\Vert \tag{1}$$
Proof: Let $[x,y]$ be a segment in $A$.
If $f(y)=f(x)$, then $(1)$ obviously holds.
Let $f(y) \not = f(x)$, then let $v_0 := \frac{f(y)-f(x)}{\Vert f(y)-f(x)\Vert} \in \mathbb R^m.$ We define function $\varphi : [0,1]\to A$, as $\varphi(t) := (1-t)x+ty.$ Definition is valid since [x,y] is in $A$. We also define $s : \mathbb R^m \to \mathbb R, s(v) := (v_0|v).$
Now we see that $D\varphi(t)=y-x$ and $Ds(d)v = (Dv_0(d)v|I_m(v)) +(DI_m(d)v|v_0)=0 +(I_mv|v_0)=(v_0|v),\forall d,v\in \mathbb R^m.$
Now lets look at the composition $g : [0,1]\to\mathbb R, g:=s\circ f\circ \varphi$ Which is continous on $[0,1]$ and differentiable on $\langle0,1\rangle$. From mean value theorem, there exists $t \in \langle 0,1\rangle$, such that $g(1)-g(0)=g'(t)(1-0)=g'(t)$. $$g'(t)=D(s\circ f\circ \varphi)(t)=Ds(f(\varphi(t)))Df(\varphi(t)D\varphi(t)\\=(v_0|Df((1-t)x+tx)(y-x))$$
On the other hand $g(1) - g(0) = s(f(\varphi(1)))-s(f(\varphi(0)))=(v_0|f(y))-(v_0|f(x))=(v_0|f(y)-f(x))=\Vert f(y)-f(x)\Vert$.
So we have: $$\Vert f(y) - f(x)\Vert = (v_0|Df((1-t)x+tx)(y-x)) \\\le \Vert v_0\Vert \cdot \Vert Df((1-t)x+ty)(y-x)\Vert=\Vert Df((1-t)x+ty)(y-x)\Vert\\\le \Vert Df((1-t)x+ty) \Vert_o\cdot\Vert(y-x)\Vert\le\Vert Df\Vert_\infty\cdot\Vert y-x\Vert$$.
My question is, is this proof valid and if it is, did i sneak somewhere that all partial derivatives of $f$ are continous on $A$ since the proof in my book assumes that and this one doesn't.