Let $A = \lbrace (tr,1t)\;  \; t \in [0,1], r \in \Bbb{Q}\rbrace$. Is it true that any continuous function from $A$ into $A$ has a fixed point?

Yes (I assume $A$ has the induced topology). The point $V:=(0,1)$, common endpoint of all segments $S_r:=\{(rt,1t): t\in[0,1]\}$, is either fixed by the continuous function $f:A\to A$, or it is mapped into some $S_r\setminus\{V\}$. But then $f $ has a fixed point on $S_r$. 

