Fixed-point property and $T_0$ separation property Each topological space $A$ with fixed-point property is $T_0$ space. Proof: suppose, two different points $a_1$ and $a_2$ belong to the same open subsets of $A$. Then the function
$$f(a)=
\begin{cases}
a_1 \quad if\,\, a=a_2\\
a_2 \quad if\,\,a\neq a_2
\end{cases}
$$
is a continuous function without fixed points. Is there an intuitionistic analogue of this result?
 A: Consider in intuitionistic mathematics the example of $A$ being the unit interval $[0,1]$ with the trivial topology $\{\emptyset,[0,1]\}$.
Then $A$ is not $T_0$ and yet $A$ still has the fixed-point property since any real function has to be continuous in the usual topology. The fixed-point property here should be read intuitionistically, that is: for a continuous function $f$ it is impossible that $f(x)\# x$ for all $x\in A$.
So I doubt there is a good intuitionistic analogue. The moral here is that intuitionistic spaces have a natural topology (the apartness topology, which for Polish spaces coincides with the metric topology) which cannot be circumvented. This natural topology however is always $T_1$...so there is no big need for an intuitionistic analogue of the above statement.
There is more on intuitionistic separation properties $T_0$-$T_4$ in my PhD thesis modern intuitionistic topology.
[with thanks to George Cherevichenko for pointing out the glaring error in my first mistaken answer]
