# Bounded Jacobian implies uniform continuity

I am trying to solve the following problems but I am not sure what the difference between the 2 problems is.

1) Prove that is $U = B_r(x)$ (open ball centered at $x$ with radius $r>0$) is an open subset of $\mathbb{R^n}$ and $f:U\rightarrow \mathbb{R^m}$ is a continuously differentiable function with uniformly bounded Jacobian (there is a constant $K < \infty$ such that $||Df(x)|| \leq K$ for all $x\in U$) then $f$ is uniformly continuous.

2) Same problem as before but this time $U$ is just an open path-wise connected subset of $\mathbb{R^n}$. Is $f$ in this case uniformly continuous?

I cannot see what is the key difference between the 2 problems and why the second case might not work.

I would very much appreciate some help. Thank you

The key difference between the two situations is that in the first situation, you know that when two points $x,y \in U$ are close, the straight line segment connecting the two points lies entirely in $U$, and that allows you to give a bound on $\lVert f(x)-f(y)\rVert$ that shows the uniform continuity (the Lipschitz continuity even).
In the second situation, two points may be close to each other but there is no short path between the two in $U$. Then the argument of the first situation breaks down. Analyse where exactly it breaks down, and use that to construct a counterexample.