Showing a contraction without a fixed point

Use $$ f(x)-f(y) = x - y + \frac{1}{x} - \frac{1}{y} = (x-y) \left(1 - \frac{1}{x y}\right) $$ Thus, for $x\not= y$ $$ \left| f(x)-f(y) \right| = \left|x-y\right| \left(1 - \frac{1}{x y}\right) < \left|x-y\right| $$ since $1 - \frac{1}{x y} < 1$ for all $x>1$ and $y>1$

Look at user40314's answer for the inequality. Regarding your second question, the space $[1, \infty)$ is complete although your map isn't a contraction map because for that, you would need there to be a constant $r$ outside the right hand side of the inequality for which $|r| < 1$. In your case, there is a factor of $1 - \frac{1}{xy}$ which can be made arbitrarily close to $1$ by picking large enough $x$ and $y$.

I want to extend kahen's comment.

It is in fact related to the contraction definition: a map such that $$d(f(x), f(y)) < d(x,y)$$ is called contractive, whereas one such that there exists some $0<k<1$ such that $$d(f(x), f(y)) \leq kd(x,y)$$ $\forall x, y$ is called a contraction (which is required in Banach's theorem). [Note that contraction $\implies$ contractive but that the opposite is wrong]

The given function (where $d$ is given by $d(x, y) = |x-y|$) is a good counterexample showing that one can not relax the contraction condition in Banach's theorem to contractive. [Note that there is, however, a theorem by M. EDELSTEIN giving a sufficent condition on a contractive map to admit a fixed point]