Given an open set $U \subset \mathbb{R}^{m}$ connected by paths, consider the metric in $U$ given by $$d(x,y) = \inf_{\gamma \in C(x,y)}\int_{0}^{1}\Vert \gamma'(t) \Vert\text{d}t$$ where $C(x,y)$ is the set of the curves $C^{1}$ by parts $\gamma: [0,1] \to U$ such that $\gamma(0) = x$ and $\gamma(1) = y$. Show that if $f: U \to \mathbb{R}^{n}$ is differentiable and $\Vert df(x) \Vert \leq K$ for any $x \in U$ then $\Vert f(x) - f(y) \Vert \leq Kd(x,y)$ for any $x$ and $y$ in $U$.
My attempt. $$\|f(x) - f(y)\| = \left\|\int_{x}^{y}f'(t)\text{d}t \right\| = \left\|\int_{0}^{1}f'(\gamma(t))\|\gamma'(t)\|\text{d}t \right\| \leq \int_{0}^{1}\|f'(\gamma(t))\|\|\gamma'(t)\|\text{d}t \leq K\int_{0}^{1}\|\gamma'(t)\|\text{d}t.$$ Now, since $\gamma$ is arbitrary, we can choose $\gamma$ such that $\int_{0}^{1}\|\gamma'(t)\|\text{d}t = d(x,y)$?
My solution seems too simple. I suspect something is wrong.
Let first $\gamma\colon[0,1]\to U$ be continuously differentiable. Then one may choose some $\delta>0$ such that for all $t_0=0<t_1<\ldots<t_{n-1}<t_n=1$ with $t_{i+1}-t_i<\delta$ there is some $r>0$ such that the ball $B(\gamma(t_i),r)$ is a subset of $U$ containing $\gamma([t_i,t_{i+1}])$. Then $\Vert f(\gamma(t_{i+1}))-f(\gamma(t_i))\Vert\leq K\Vert\gamma(t_{i+1})-\gamma(t_i)\Vert$ for all $i$ since $B(\gamma(t_i),r)$ is convex. Thus $\Vert f(\gamma(1))-f(\gamma(0))\Vert\leq K\sum_{i=0}^{n-1}\Vert\gamma(t_{i+1})-\gamma(t_i)\Vert$. Taking the supremum of the LHS we get $\Vert f(\gamma(1))-f(\gamma(0))\Vert\leq K \cdot\text{length}(\gamma)$, where the length of $\gamma$ is the supremum of the lengths of all inscribed polygons which is the same as $\int_{0}^{1}\|\gamma'(t)\|\text{d}t$. If $\gamma$ is piecewise continuously differentiable one may apply this result to the corresponding pieces.
Thus you get $\Vert f(\gamma(1))-f(\gamma(0))\Vert\leq K \int_{0}^{1}\|\gamma'(t)\|\text{d}t$ for all $\gamma\in C(x,y)$. Taking the infimum over $\gamma$ finishes the argument.