Rather than speculate about the importance of the difference between inhabited and non-empty, I will simply explain what the difference is in the context of topos theory.
Definition. An inhabited object is an object $X$ for which the sentence $(\exists x \in X . \, \top)$ is true. A non-empty object is an object $X$ for which the sentence $\lnot (\forall x \in X . \, \bot)$ is true.
Here, of course, I am interpreting formulae using the Kripke–Joyal semantics internal to a topos. (Set theorists may wish to substitute "intuitionistic Kripke model of IZF" for "topos".) The formulae I have used are not exactly the same as Goldblatt's, because Kripke–Joyal semantics can only interpret bounded quantifiers; if you think about it a little you will see that the usual translation of bounded quantifiers to unbounded quantifiers recovers Goldblatt's version.
What does this mean in the external logic? Well, an object $X$ is inhabited if and only if the unique morphism $X \to 1$ is an epimorphism, and $X$ is non-empty if and only if, for all subobjects $U \subseteq 1$, if $X \times U$ is empty, then $U$ itself is empty.
For concreteness, let $\mathcal{E} = \textbf{Sh}(B)$ be the topos of sheaves on a non-trivial topological space $B$, say, the circle $S^1$. An object of $\mathcal{E}$ is a sheaf on $B$, i.e. a topological space $E$ equipped with a local homeomorphism $E \to B$, and a morphism in $\mathcal{E}$ is a continuous map making the obvious triangle commute. It is not hard to find examples of inhabited objects: the terminal sheaf $1$, corresponding to the identity map $\textrm{id} : B \to B$, is inhabited for trivial reasons. A more interesting example would be the 2-to-1 connected covering space $E \to B$ of the circle: this is inhabited, but it is also impossible to find a "global" element of $E$.
In general, any inhabited object is non-empty, but here the converse is false: if we take $V$ to be a dense open subset of $B$, say, $B$ minus a point, then the sheaf $V \hookrightarrow B$ will be non-empty but not inhabited. Why is this? Well, clearly, $V \hookrightarrow B$ is not an epimorphism of sheaves, and the only open subset $U \subseteq B$ such that $U \cap V = \emptyset$ is $U = \emptyset$ itself. More generally, a sheaf $E$ on $B$ will be non-empty if and only if the image of the projection map $E \to B$ is dense in $B$. (Set theorists familiar with forcing should find this vaguely familiar: this is precisely the interpretation of existence needed to make a model satisfy classical logic.)