I'm having a hard time directly proving this without referring to any calculus based theorems on limits on big-$O$, $\omega$, and $\theta$.
Since I am disproving I will want to show $ \neg(n \log n \in \Omega (n^2))$ I can't directly figure out a way to set a value for n such that I can prove $ \neg(n \log n \in \Omega (n^2))$.
For example, since the constant c can be a real number greater than $0$ but less than $1$, makes it more difficult in finding an n that satisfies all $C$ values.
So suppose for a contradiction that $n\log n \in \Omega(n^2)$.
So by definition $$\exists C>0: \exists n_0 \in \Bbb N: \forall n \ge n_0: n \log n \ge Cn^2\tag{1}$$
dividing by $n$ on both sides of the inequality gives
$$\exists C>0: \exists n_0 \in \Bbb N: \forall n \ge n_0: \log n \ge Cn \tag{2}$$
Now enter calculus (yes you do need it, in maths you often build on facts that you proved before, or what's the point otherwise?):
$$\lim_{n \to \infty} \frac{\log n}{n}=0\tag{3}$$
Recall that a limit of a sequence (of reals) is defined as
$$\lim a_n = L \iff \forall \varepsilon>0: \exists N \in \Bbb N: \forall n \ge N: |a_n - L| < \varepsilon$$
and apply that to $a_n = \frac{\log n}{n}$, $L=0$ and $\varepsilon = \frac{C}{2}>0$ and we get $N \in \Bbb N$ such that $$\forall n \ge N: \frac{\log n}{n} < \frac{C}{2}\tag{4}$$
and so taking $n_1 = \max(N, n_0)$ ($n_0$ from $(1)$) and multiplting the previos by $n$ we get, as $n_1 \ge N$ from $(4)$:
$$\log n_1 < \frac{C}{2} \cdot n_1$$
and from $(2)$, as $n_1 \ge n_0$ that
$$\log n_1 \ge Cn_1$$
and it's clear that those last two equations together form a blatant contradiction.
So the assumption $n\log n \in \Omega(n^2)$ has led to a contradiction and is false.