Verify that dominance is a partial ordering.
Let C be an ascending chain of dominant local rings, and let U be their union. Two nonunits in U are nonunits in some ring R in the chain. Remember that R is a local ring. Let z be their sum. If z is a unit in U then 1/z appears in some ring S in the chain. Now z and 1/z are units in S. Intersect S and R, and if z is common to both, and a nonunit in R, it should be a nonunit in S. This is a contradiction, hence z is a nonunit in U. The sum of nonunits remains a nonunit.
Similarly we can multiply a nonunit in U by anything else in U and obtain a nonunit, so the nonunits form an ideal, and U is local.
Intersect U with any ring R in the chain, and the nonunits remain nonunits. Therefore U dominates the entire chain. We have found a local ring that dominates C. Every ascending chain of local rings is bounded by a local ring, i.e. the union, that dominates the entire chain. By zorn's lemma there is always a maximal local ring in F, that dominates any given local ring R, though this maximal ring may be F itself.
Such is the case when F is the union of all finite fields of characteristic p. If R is a ring inside F, and x is an element of R, then Zp adjoin x creates a finite field in R, every x is invertible, and R is a field. All the rings in F are fields.
Every subring of F is either F itself, or a finite field. Fields are ordered by containment, and each is technically a local ring, with 0 acting as the maximal ideal.
Now start with any finite field K in F and build a chain of ascending finite fields, a chain of local rings, each dominating the previous. The "maximal" local ring is the entire field F.
We are going to prove that a maximal local ring (with respect to dominance) is a valuation ring, but first we need a lemma.