Let $p: X\to Y$ be a quotient map of topological spaces. Is it true that for all $z \in Y$, there is an open neighbourhood $U\subset Y$ with $z\in U$ and a section $s: U\to X$, so that $(p\circ s)(y) = y$ for all $y\in U$?
I am happy to assume some hypotheses to exclude pathological examples.
Counterexample: Let $X = ([-1, 0] \times \{1\}) \cup ([0, 1] \times \{2\}) \subset \mathbb{R}^2$ be the union of the closed line segments from $(-1, 1)$ to $(0, 1)$ and from $(0, 2)$ to $(1, 2)$; let $Y = [-1, 1] \subset \mathbb{R}$; let $p(x, y) = x$. Then any possible map $s$ is discontinuous at $0$.