I have produced a false proof but can't spot the mistake.
I proved the following (false) statement:
Let $U \subseteq \mathbb R^n$ be open and $f: U \to \mathbb R^n$ be continuous. Then $f$ is locally Lipschitz.
Note that this statement is false as continuity does not imply local Lipschitzness.
Please could someone tell me where I my mistake is?
Here is my false proof:
The goal is to show that for a point $u \in U$ there is an open set containing $u$ on which $f$ is Lipschitz. So let's pick some $u$ in $U$. Since $U$ is open there exists some open ball $B = B(u,r)$ such that the closure of $B$ is also contained in $U$:
$$ u \in \overline{B} \subseteq U$$
Since $f$ is continuous and $\overline{B}$ is compact, $f$ is uniformly continuous on $\overline{B}$. That is, for every $L > 0$ there is a $\delta>0$ such that
$$ |x-y|<\delta \implies |f(x) - f(y)|<L$$
Fix $L=1$ (an arbitrary choice!) and pick $\delta' $ to be smaller than $\min(r, \delta,1)$. Then $B(u, \delta')$ is contained in $B(u,r)$.
Then $f$ is Lipschitz continuous with Lipschitz constant $L$ on $B(u,\delta')$:
If $x,y \in B(u, \delta')$ then $\|f(x) - f(y)\| < 1 < 1\cdot \|x-y\| = L \|x-y\|$.
Your argument is correct up to $\|f(x) - f(y)\| < 1$. It fails at the next step, as the inequality $1 < 1\cdot \|x-y\|$ is false in general.
Getting the factor of $\|x-y\|$ in this estimate is the essence of Lipschitz continuity. And you don't have it.