Let $X$ be an algebraic stack. Given a point $f:T\to X$, we define $Stab(f)=X\times_{X\times X}T$, where the map $X\to X\times X$ is the diagonal and $T\to X\times X$ is $(f,f)$. We say that the stabilizer if affine if $Stab(f)\to T$ is an affine morphism.

Since affine morphisms are stable under base extension, it is always true that if the diagonal is affine, then stabilizers are affine.

If I've got everything right so far, then I think I have an argument that shows that if stabilizers are affine, then the diagonal is affine. Let $h:U\to X$ be a smooth cover by an affine scheme, then $Stab(h)$ is affine over $U\times U$ by assumption. But affine morphisms are local on the base in the smooth topology, so the diagonal is affine.

    Stab(h) --> U×U
      |          | 
      |  cart    | smooth cover
      v          v
      X ------> X×X

But this uses that stabilizers of *scheme-theoretic* points are affine. Perhaps in the original question, you're only allowed to assume that stabilizers of geometric points are affine (or something like that).

**Edit:** Somehow I missed that "closed points" part of the question.