Define $f:[1,\infty)$ by $f(x)=x+\frac{1}{x}$. Now, we have the following:
(1) This space is complete
(2) For any $x,y\in[1,\infty)$ we have $|f(x)-f(y)|=\left|\left(x+\frac{1}{x}\right)-\left(y+\frac{1}{y}\right)\right|=\left|x-y+\left(\frac{1}{x}-\frac{1}{y}\right)\right|\le|x-y|+\left|\frac{1}{x}-\frac{1}{y}\right|<|x-y|$, by the triangle inequality. Therefore for any $x,y\in[1,\infty)$ we have $|f(x)-f(y)|<|x-y|$, and so this mapping is a strict contraction.
(3) $f$ has no fixed point - note that if $g(x)=x$ then $x+\frac{1}{x}=x$ then $\frac{1}{x}=0$, which is impossible.
The question is: How does this not contradict the Contraction Mapping Theorem?
The Contraction Mapping Theorem says that for any complete metric space, if a contraction is strict then the function has exactly one fixed point. But I cannot see which part of the definition fails in this case.
This mapping isn't a strict contraction (even after you fix your proof of the inequality). At least, not by the standards of the requirements of the theorem.
The contraction mapping theorem requires that there is some $q \in (0,1)$ such that $|f(x) - f(y)| \leq q|x-y|$. In other words, not only does $f$ contract the space but there is some scale factor that the contraction always does at least as well as.
Your $f$ contracts, but as $x$ and $y$ get larger and larger the scale factor gets closer and closer to $1$. There is no $q$ that will work on the entire space. This sort of function is exactly what is being excluded by the theorem's requirement of a strict contraction.
(One useful exercise would be to go through the proof of the contraction mapping theorem again, see where the $q$ is used, and see how the proof will fail for your $f$)