Simply connected does not imply contractible. Is there a nice counter example in $R^2$?

The standard counter example to the claim that a simply connected space might be contractible is a sphere $S^n$, with $n > 1$, which is simply connected but not contractible. Suppose that I were interested in a counter example in the plane - does anyone know of a subset of $R^2$ which is simply connected but not contractible?

Consider the topologist sine curve

$$y = \sin \bigg(\frac{1}{x}\bigg),\ 1\geq x>0$$

together with the interval $\{(0, t): |t|\leq 1\}$ and a curve joining this interval with the graph. This is simply connected but is not contractible. You may find the proof of noncontractibility in

http://math.ucr.edu/~res/math205B-2012/polishcircle.pdf

John's answer is absolutely correct, but here is an addendum to it: All proper subsets $Z$ of the plane are aspherical in the sense that every continuous map $$ f: S^n\to Z, n>1, $$ extends to a continuous map of the ball, $B^{n+1}\to Z$. See the paper One-dimensional sets and planar sets are aspherical by Cannon, Conner and Zastrow.