$(X,d)$ is a complete metric space. $\liminf_{n \to \infty} \sup_{x,y \in X; x \neq y} \frac{d(f^n(x),f^n(y)}{d(x,y)}<1$ then $f$ has a fixed point.
From the statement $\liminf_{n \to \infty} \sup_{x,y \in X; x \neq y} \frac{d(f^n(x),f^n(y)}{d(x,y)}<1$. I have proved that(after a long calculation) there exists $r \in \Bbb N$ s.t
$$d(f^r(x),f^r(y))\leq q d(x,y)$$ for some $q \in (0,1)$[I don't know if we can get more general result or not!]. Using Banach Fixed point theorem I got that $f^r$ has a fixed point. How do I conclude for $f$?
It is not difficult to see that $\liminf_{n \to \infty} \sup_{x,y \in X; x \neq y} \frac{d(f^n(x),f^n(y)}{d(x,y)}<1$ is in fact equivalent to the fact that there exists an $r\in\mathbb{N}$ such that $$d(f^r(x),f^r(y))\leq q d(x,y)$$ for some $q \in (0,1)$, i.e. $f^r$ is a contraction.
Now Banach theorem is in fact stronger than what you've said, in that it states that the fixed point of $f^r$ is unique. Let $p$ be the fixed point of $f^r$ and observe that $f(p)$ is also a fixed point of $f$: $$f^r(f(p)) = f(f^r(p)) = f(p)$$ (if you want a more geometric proof of it, draw the orbit of point $p$ under the action of $f$ - $f$ acts as a cyclic group on the orbit of $p$)
But the fixed point is unique, so we must have $f(p) =p$
I will also add, that the fact that $f^r$ is a contraction does not imply that $f$ is - as a counterexample, you can take the linear map $$A=\begin{pmatrix} 0 & 1\\ 0 & 0 \end{pmatrix} $$ we have $A^2 = 0$ which is clearly a contraction, but on the other hand $$A\begin{pmatrix}0 \\ 1\end{pmatrix}=\begin{pmatrix}1 \\ 0\end{pmatrix}$$ so $A$ is not contractive in the standard Euclidean metric.