Converse to Banach's fixed point theorem? Let $(X,d)$ be a metric space. Banach's fixed point theorem states that if $X$ is complete, then every contraction map $f:X\to X$ has a unique fixed point. A contraction map is a continuous map for which there is an real number $0\leq r < 1$ such that $d(f(x),f(y))\leq rd(x,y)$ holds for all $x,y\in X$.

Suppose $X$ is a metric space such that every contraction map $f:X\to X$ has a unique fixed point. Is $X$ complete?

 A: As has been pointed out, the putative converse to the Contraction Mapping Theorem suggested in the question is not true.  But there is a result which may reasonably be viewed as the converse of CMT.  
Theorem (Bessaga, 1959): Let $X$ a set and $f: X \rightarrow X$ a function such that for all $n \in \mathbb{Z}^+$, the iterate $f^n$ has a unique fixed point.  Then there exists a complete metric on $X$ with respect to which $f$ is a contraction mapping (for any preassigned constant $c \in (0,1)$).  
[Addendum: I just looked at Elekes' paper and saw that it cites this result of Bessaga and says that it is seemingly the earliest converse.  So I guess this post is not exactly exciting news.  Oh well.]  
I learned about this result from a talk that Keith Conrad gave in the Undergraduate Math Club at UGA.  For more information, see his "blurb"
http://www.math.uconn.edu/~kconrad/blurbs/analysis/contractionshort.pdf
In later correspondence with him I pointed out the following result, which is now included in his writeup:
Theorem: For a function $f: X \rightarrow X$, the following are equivalent:
(i) Every iterate $f^n$ has at most one fixed point.
(ii) There is a metric (not necessarily complete!) with respect to which $f$ is a contraction.
This is not an earth-shattering result but it has a nice, crisp statement and afterwards I decided that it was too good to be true that I was the first to think of it.  And I was right -- after a quick internet search I found the result in a published paper.  (Unfortunately I didn't take note of the reference.  Sorry, K. EDIT: It was the paper Jacek Jachymski: A short proof of the converse to the contraction principle and some related results, Topological Methods in Nonlinear Analysis, Volume 15, Number 1 (2000), 179-186; DOI: 10.12775/TMNA.2000.014, projecteuclid.)
A: If you want a sort of positive result, this comes to my mind, for what is worth:  

If $X$ is a metric space such that any
  contraction map $T:Y \to Y$ on any nonempty
  closed subset of X has a fixed point,
  then $X$ is complete.

Indeed, a Cauchy sequence $(x_n)$ converges if and only if some subsequence converges. Up to extracting a subsequence, a Cauchy sequence is either stationary (thus convergent) or injective, and verifies $d(x_{p+1},x_{q+1})\leq d(x_p,x_q)/2$. 
If the set $\{x_n: n\in \mathbb{N}\}$ were closed in $X$, $T(x_n):=x_{n+1}$ defines a contraction with no fixed points there. Hence it's not closed, thus it's closure contains exactly one more point, the limit of the sequence.
A: The answer is no, for example look at the graph of $\sin(1/x)$ on $(0,1]$. But for more information and related questions check out "On a converse to Banach's Fixed Point Theorem" by Márton Elekes.
A: 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 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.pdf
A: Yes if every Kannan mapping $f$ ,$\{f$ the Kannan if there exists $h\in[0,\frac{1}{2})$ such that $d(f(x),f(y))\leq h\{d(x,f(x))+d(y,f(y))\}$\, for all $x,y\}$ has a fixed point .
