Here is a negative answer for your additional remark concerning Banach's fixed point theorem: Consider $$d(x,y)=(\int_0^1|x-y|^p)^{1/p}$$, $0<p<1$ which satisfies the quasi-triangle inequality. Look at the set of all (measurable real) functions on $[0,1]$ boundaed between 0 and 2 and of integral 1. Look at the Baker transformation on this set: first map $x$ to $y(t)=2x(2t)$, $0\le t\le 1/2$ then trancate at hight 2 and shift what remains ($(y-1)^{+}$) by $1/2$ to the right and $2$ down. I think I checked that it is a contraction (with constant $2^{p-1}$). This map is known not to have a fixed point, see the following paper of Dale Alspach: http://www.claremontmckenna.edu/math/moneill/Math%20138/papers138/Alspach.pdf

Here is a negative answer for your additional remark concerning Banach's fixed point theorem: Consider $d(x,y)=(\int_0^1|x-y|^p)^{1/p},$ $0<p<1$ which satisfies the quasi-triangle inequality. Look at the set of all (measurable real) functions on $[0,1]$ boundaed between 0 and 2 and of integral 1. Look at the Baker transformation on this set: first map $x$ to $y(t)=2x(2t)$, $0\le t\le 1/2$ then trancate at hight 2 and shift what remains ($(y-1)^{+}$) by $1/2$ to the right and $2$ down. I think I checked that it is a contraction (with constant $2^{p-1}$). This map is known not to have a fixed point, see the following paper of Dale Alspach: http://www.claremontmckenna.edu/math/moneill/Math%20138/papers138/Alspach.pdf