Banach Fixed Point Theorem states that if \((X,d)\) is a complete metric space and \(f\colon X\to X\) is self map such that there is a constant \(q\in \[0,1)\) with the condition \[ d(f(x),f(y)) < q \cdot d(x,y) \] then \(f\) has a fixed point, i.e. there is a point \(a\in X\) such that \(f(a)=a\).
Assume \((X,d)\) is a compact metric space and \(f\colon X\to X\) is a continuous self-map such that \[d(f(x),f(y)) < d(x,y)\] i.e. \(f\) is a weak contraction. Then \(f\) has a fixed point, i.e. there is a point \(a\in X\) such that \(f(a)=a\).
Our condition is weaker on the inequality. However, we require that \(X\) is compact, which is stronger than complete.
First, define a map \(g\colon X\to \mathbb{R}\) by letting \[g(x) = d(f(x),x)\] Then \[ g(x) = d(f(x),x) \leq d(f(x),f(y)) + d(f(y),y) + d(y,x) = d(f(x),f(y)) + d(x,y) + g(y) < 2 d(x,y) + g(y) \] and \[ g(y) = d(f(y),y) \leq d(f(y),f(x)) + d(f(x),x) + d(x,y) = d(f(x),f(y)) + d(x,y) + g(x) < 2 d(x,y) + g(x) \] which means \[ \|g(x)-g(y)\| < 2 d(x,y) \] i.e. \(g\) is (uniformly) continuous. Since \(X\) is compact, \(g\) achieves its minimum and maximum on \(X\). Let \(a\in X\) be such that \[ g(a) = d(f(a),a) \leq d(f(x),x) = g(x) \] for all \(x\in X\). But then \[ g(f(a)) = d(f(f(a)),f(a)) < d(f(a),a) = g(a) \] is a contradiction unless \(f(a) = a\), i.e. \(f(a)=a\) is a fixed point.
The answer is no! If we relax the contraction condition and assume \(d(f(x),f(y))\leq d(x,y)\) the statement fails: take \(X = \{-1,1\}\) and \(f(x)=-x\) then \(f\) is and isometry but \(f\) has no fixed point. On the other hand, if we relax the condition on compactness the statement again fails: take \(X=\{\frac{1}{n}\|\ n=1,2,\ldots\}\) with the ambient metric in \(\mathbb{R}\). Define \(f(x) = \frac{x}{x+1}\), then \(f(x)\) satifies \(d(f(x),f(y)) < d(x,y)\) for every \(x,y\in X\) and there is no fixed point.