Let $(X, d)$ be a metric space and $f\colon X \to \mathbb{R}$ be lipschitz on open sets $U_{1}, U_{2}\subseteq X$. Prove or disprove that $f$ is lipschitz on $U_{1} \cup U_{2}$.
I first tried to prove it: Fix some $u_1 \in U_{1}$ and $u_2 \in U_{2}$ and $M \in \mathbb{R}$ s.t. \begin{align*} \left| f( u_{1} ) - f( u_{2}) \right| \leqslant M \cdot d( u_{1}, u_{2}). \end{align*} Now, for any two points $a \in U_{1}, \; b \in U_{2}$ we have \begin{align*} \left| f( a) - f( b) \right| &= \left| f( a) - f( u_{1}) + f( u_{1}) - f( b) \right| \\ &\leqslant \left| f( a) - f( u_{1}) \right| +\left| f( u_{1}) - f( u_{2}) + f( u_{2}) - f( b) \right| \\ &\leqslant \left| f( a) - f( u_{1}) \right| +\left| f( u_{1}) - f( u_{2}) \right|+ \left| f( u_{2}) - f( b) \right| \\ & \leqslant L_{1}d( a, u_{1}) + M d( u_1, u_2) + L_{2}d( u_{2}, b) .\end{align*} At this point I don't know how to continue, so I'm starting to doubt whether this is actually true.
Take $X=\{1/n:n\in\mathbb N\}$, with the standard metric of the real numbers.
Then $U_1=\{1/(2n):n\in\mathbb N\}$ and $U_2=\{1/(2n+1):n\in\mathbb N\}$ are open subsets of $X$.
Set $f(1/n)=0$ if $n$ odd, and $f(1/n)=1$ if $n$ is even.
Then $f$ is Lipschitz when restricted on either $U_i$ but discontinuous!