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This is now an old thread, but I just came across it. The question you are asking was asked by Behrends a couple of years ago, and he knew about the counter example on the plane that Gjergji points out. Behrends was specifically asking for an incomplete metric space that is a subset of the line that and has the fixed point property.

I came up with a partition of $\mathbb R$ into two dense sets $X_0$ and $X_1$ such that for all $i\in{0,1}$, every map $f:X_i\to X_i$ that satisfies $$\forall x,y\in X_i(x\not=y\rightarrow|f(x)-f(y)|<|x-y|)$$ is actually constant (and in particular has a fixed point). The $X_i$ can be constructed by an easy transfinite recursion and there is no control of how complicated the sets are. This is the huge advantage of Elekes' examples which are Borel.

The notes with my proof are here: http://www.hausdorff-center.uni-bonn.de/people/geschke/papers/Behrends.dvihttp://www.hausdorff-center.uni-bonn.de/people/geschke/papers/Behrends.pdf

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This is now an old thread, but I just came across it. The question you are asking was asked by Behrends a couple of years ago, and he knew about the counter example on the plane that Gjergji points out. Behrends was specifically asking for an incomplete metric space that is a subset of the line that has the fixed point property.

I came up with a partition of $\mathbb R$ into two dense sets $X_0$ and $X_1$ such that for all $i\in{0,1}$, every map $f:X_i\to X_i$ that satisfies $$\forall x,y\in X_i(x\not=y\rightarrow|f(x)-f(y)|<|x-y|)$$ is actually constant (and in particular has a fixed point). The $X_i$ can be constructed by an easy transfinite recursion and there is no control of how complicated the sets are. This is the huge advantage of Elekes' examples which are Borel.

The notes with my proof are here: http://www.hausdorff-center.uni-bonn.de/people/geschke/papers/Behrends.dvi
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