Counterexample of Metric Spaces

Question

I know the result that if X and Y are 2 metric spaces with Y complete and f is uniformly continuous on a dense subset D of X then f can be continuously extended to X.
Can someone show that this result fails if Y is not complete?
Thanks for any help.

Answer 1

How about $D=[0,1) \subset [0,1]=X,Y=[0,1)$ and $f(x)=x$?