Here's a proof which worked for both orientable and non-orientable manifolds. The proof comes from (my memory of) Milnor's book "Topology from the Differentiable Viewpoint". The proof only shows there is no smooth retraction. (With more work, one can show that there is a continuous retraction iff there is a smooth one, so this can be turned into a proof of the general statement.)
Suppose for a contradiction that $r:M\rightarrow \partial M$ is a retraction of a compact manifold $M$ onto its boundary. By Sard's theorem, there is a regular value $p\in \partial M$.
Then $r^{-1}(p)$ is a $1$-d submanifold in $M$. It is closed, being the inverse image of the closed set $\{p\}$, so is compact. A compact $1$-d manifold is homeomorphic to a disjoint union of a finite number of circles and closed, bounded intervals. In particular, the boundary of $r^{-1}(p)$ has even cardinality.
Now, the key point is that the boundary of $r^{-1}(p)$ must lie in $\partial M$. But $r^{-1}(p)\cap \partial M = \{p\}$ (since $r$ is a retraction), so is not of even cardinality.
Best Answer
It seems that $[0;\infty)$ is the required example.