Counterexample to a "modified" Banach Fixed Point Theorem?

Question

The Banach theorem states that if a (self) map on a complete metric space is Lipschitz with ratio $< 1$, it has a unique fixed point. What about modifying the hypotheses to say that the only condition the map satisfies is that $d(x, y) > d(f(x), f(y))$ where $x, y$ are in the domain? (I think this condition is known as "weakly contracting" I think that this type of map would fix a point if the metric space were compact, but what about in the general case where the metric space is not compact? Is there a counterexample?

Answer 1

$X=[1,\infty)$ with the usual metrix, $f(x)=x+e^{-x}$. Use Mean Value Theorem to verify that the hypothesis is satisfied.

Answer 2

$$ \frac{3x + \sqrt {x^2 + 1}}{4} $$