This question was prompted by this one: If $\Delta \vDash \exists x.p(x)$, then $\Delta \vDash p(\tau)$ for some ground term $\tau$. Why is this false?, but is a proof-theoretic inquiry into why does the reverse of Existential Generalisation not hold (which I have the intuitive feeling is related to the statement below).
The statement in the aforementioned question was as follows:
If $\Delta \vDash \exists x.p(x)$, then $\Delta \vDash p(\tau)$ for some ground term $\tau$.
I didn't quite get the answers given because I am not well versed in model theory or anything beyond standard n-order logic. I get why a specific p(a) or p(b) or whatever is not entailed, but why is it that no ground term p(τ) is entailed? And, given that many have pointed out that no ground term is entailed in the special case there is no ground term at all, why does the entailment hold in case where we have universal quantification instead?