Let $K$ be a number field or finite field. Let $C$ be a singular curve defined over $K$. Let $C'$ be normalization of $C$ and $\phi : C'\to C$ be normalization map.
It is well known that this map $\phi$ is surjective (for example, Stacks 035E), and $\phi $ is defined over $K$. Then, does $C(K)\neq \emptyset$ imply $C'(K)\neq \emptyset$・・・① ?
But in this question, Sasha reads we can construct counterexample of ①.
So I'm confused. Why ① does not hold in spite of $\phi$ is surjective and defined over $K$ ?
No, this does not hold. Here is a counterexample: let $C=\operatorname{Spec} \Bbb Q[x,y]/(x^2+y^2)$, which has normalization $C'=\operatorname{Spec} \Bbb Q[t,u]/(u^2+1)$. Then $C(\Bbb Q)=\{(x,y)\}$ while $C'(\Bbb Q)=\varnothing$.
The reason your desired claim doesn't hold is that if $p\mapsto q$ for a map of schemes $f:X\to Y$, we have a map of residue fields $k(q)\to k(p)$. This means "size limitations" only go one way - if $k(p)$ is "small", then $k(q)$ must be as well; if $k(q)$ is "big" then $k(p)$ must be too (being a $k$-rational point of a $k$-scheme is in some sense the most strict "smallness" limitation). For instance, if $X$ and $Y$ are $k$-schemes, $f$ is a $k$-morphism, and $p$ is a $k$-rational point, then we have a map $k\to k(q)\to k(p)$ where the composite $k\to k(p)$ is an isomorphism, which forces $k(q)\cong k$ as well.