# For entourage $D\in \mathcal{U}$, is there $x_n\in X$ with $\operatorname{Fix}(f)=\bigcup_{n\in\mathbb{N}}D[x_n]\cap\operatorname{Fix}(f)$?

Let $$(X, \mathcal{U})$$ be a strongly Lindeloff uniform space ( it is complete separable) and $$f:X\to X$$ be a homeomorphism. A point $$x\in X$$ is called a fixed point of $$f$$ if $$f(x)=x$$. Denote by $$Fix(f)$$ the set of fixed points of $$f$$. Take $$D\in \mathcal{U}$$ a symmetric entourage and $$D[x]=\{y\in X: (x, y)\in D\}$$.

Is there a sequence $$x_n\in \operatorname{Fix}(f)$$ such that $$$$\operatorname{Fix}(f)=\bigcup_{n\in\mathbb{N}}D[x_n]\cap \operatorname{Fix}(f)?$$$$ Please help me to know it.