Completeness can be omitted from Banach Fixed Point Theorem?

"Completeness can't be omitted" is just saying that if $X$ is not complete, you can't be certain that a fixed point will exist for any given function. It doesn't mean you will never find one for specific functions.

It's like saying that "if $X$ is not complete, not all cauchy sequences converge". But you can definitely find some cauchy squences that do converge.

EDIT It's now interesting to investigate how the theorem fails without completeness:

Firstly, here's an example of a contraction defined on a non-complete metric space with no fixed point:

$f:(0,1)\rightarrow(0,1)$ such that $f(x) = \dfrac{x}{2}$.

Clearly this is a contraction because $|f(x)-f(y)|=\Big|\dfrac{x}{2}-\dfrac{y}{2}\Big| = \dfrac{1}{2}|x-y|$ for any $x, y \in (0, 1)$. However, for any $x\in(0,1)$, $f(x)\neq x$, so there is no fixed point.

(On the natural domain of the function, the fixed point would be at $0$, but you can see that this is on the boundary of our non-complete set.)

Secondly, are there any cases where the metric space being incomplete results in more than one fixed point? The answer is no. Suppose $f:X\rightarrow X$ is a contraction and $X$ is any metric space (complete or not complete). Suppose $f$ has two fixed points at $x$ and $y$. Then $|f(x)-f(y)| = |x - y|$. This contradicts the definition of a contraction which requires $|f(a)-f(b)|<|a-b|$ for all $a,b\in X$.

As well as Shanye2020's counterexample on an open interval, we can construct a counterexample on a closed subset of the rationals: take $M=[-\frac13,\frac13]\cap\mathbb Q$, and $f(x)=x^2+\frac16$. Then $f(x)$ has a fixed point in $[-\frac13,\frac13]$ as a subset of $\mathbb R$, but this fixed point is irrational.