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?



Fixed-point property and $T_0$ separation property
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](https://www.fwaaldijk.nl/modern%20intuitionistic%20topology.pdf).


[with thanks to George Cherevichenko for pointing out the glaring error in my first mistaken answer]


