In my research I need to construct a metric space $(X, d)$ with a dense subset $Y\subseteq X$ such that for every homeomorphism $f:X\to X$ we have $f(y)=y$ for all $y\in Y$. In the following I constructed such a metric space. Please help me to determine whether my idea works or still has some mistakes.
Let $A=\mathbb{Q}\cap [0, 1]$ and $$X=\cup_{x\in [0, 1]}(\cup_{a\in A}(x, a)\times (a, x)).$$ Consider $(a, a)\in X$ and a homeomorphism $f:X\to X$. Then take an sufficiently small open connected ball $U$ around $(a,a)$ such that $U\setminus\{(a, a)\}$ has four connected components. Since $f$ is a homeomorphism, hence $f(a, a)= (a, a)$.