Meaning of this- $\forall x \in S, \exists y \in S, p(x,y) \implies \exists y \in S, p(y,y)$

Question

The problem was to prove or disprove the following

$ S\neq \varnothing,\, \forall x \in S, \exists y \in S, p(x,y) \implies \exists y \in S, p(y,y)$

Answer 1

If for all x in the set S, there exists y in S such that the function p(x,y) is true, then there exists a y in S such that p(y,y) is true.

The question is basically asking if, given that every x in the set has a coresponding y that makes the statement true, if there must exist an element where the corresponding element is itself.

A counter example is easy to come up with: $S = \{1,2\}$
$p(1,2)$ is true.
$p(1,1)$ is false.
$p(2,2)$ is false.
$p(2,1)$ is true.

Every element has an element for which $p(x,y)$ is true, but no element satisfies this condition on its own.