TITLE: Lipschitz property and Lipschitz extension
QUESTION [6 upvotes]: Is there a Lipschitz function $f$ from a subset of a metric space $U$ to a complete  metric space $V$ that has no Lipschitz extension to the whole space $U$?

REPLY [3 votes]: I add two finite-dimensional examples. 

(With a topological obstruction.) Let $U=[0,1]$ (interval), $V=\{0,1\}$ (two-point set). The identity map $\{0,1\}\to \{0,1\}$ does not extend to a continuous map $U\to V$ since any continuous map $U\to V$ must be constant. 
(Without a topological obstruction.) Replace $V$ in the first example by a snowflake that contains the points $0$ and $1$. Now the identity map $\{0,1\}\to \{0,1\}$ does extend to a continuous map $U\to V$. However, there is no Lipschitz extension since any Lipschitz map $U\to V$ must be constant. (One way to prove the latter is to observe that any nontrivial subarc of the snowflake has infinite length, while the length of a Lipschitz image of $U$ must be finite.)