Showing a statement of First-Order Logic is neither a validity nor a contradiction. Working on P.D. Magnus. forallX: an Introduction to Formal Logic (pp. 267, exercise A. 2), asks:
Show that each of the following is neither a validity nor a
contradiction:
$\exists x T(x,h)$
Refute validity:
I am going to choose this interpretation:
Domain: cats, ants
$T(x,y)$: x is heavier than y.
h: cats,
a: ants
No element $d$ belonging to the domain satisfies $T(b,h)$ in interpretation $\mathbf{I}[b \setminus d]$ because
$T(h,h)$ is false, and $T(a,h)$ is false.
Is this proof correct ?
 A: As noted in the comments, there's a bit of confusion here - you seem to be interpreting $h$ as a set (namely, of all cats). However, you have the right idea: consider a domain consisting of one cat (who we name "$h$") and one ant (to whom we don't even give a specific name, nobody names ants), and interpret $T$ as "is heavier than." Then:


*

*$h$ is not heavier than itself.

*The only other element of our universe - the sadly-nameless ant - is not heavier than $h$.
So in this context, $\exists x(T(x,h))$ is indeed false. But note the shift from sets to individuals.
(Actually, we didn't even need an ant at all - just consider the domain consisting of a single cat, which isn't heavier than itself. Ants are really unnecessary in general, too.)

For showing that $\exists x(T(x,h))$ isn't a contradiction either, a similar idea should be employed - in fact, there's a couple ways to very slightly tweak the ideas above to get an answer here.
(It's separately worth giving purely mathematical example models: e.g. to show that $\exists x(T(x,y))$ is not valid, we could always work with the set $\{0\}$ and interpret $T$ as "is less than.")
