If $1\in A$ and $A$ is countably infinite then that implies there is a number $n\in\mathbb{N}$ such that $f(n)=1$. However, this would mean $f(n+1) > f(n) = 1$ and hence $f(n+1) \not\in [0,1]$, which is a contradiction. Your desired setup is therefore impossible.
Addendum: Even if you take out the condition $0,1 \in A$ the the setup is still impossible. Let $m$ be the number such that $f(m) = 1-f(1)$ - note that this is guaranteed to exist by your reflection-inclusion property. Then $f(m+1)>f(m)$ implies that $f(k) = 1-f(m+1) < 1-f(m) = f(1)$ for some likewise guaranteed-to-exist $k \in \mathbb{N}$, which contradicts $f(k)>f(1)$ (a result of the ordering property). No cake for you OP. :)
[Stated more succinctly by PZZ: the ordering implies a minimum in the set, which by reflection-inclusion implies there is a greatest element in the set, say $f(j)$, hence you get a contradiction with $f(j+1) > f(j)$.]