Could you please check my solution? It is short and simple but I doubt somehow...
Prove the following formula: $\exists x\forall u\exists y((R(x,x)\vee\neg R(x,u))\vee(R(y,x)\&\neg R(y,u)))$.
I state it is unprovable: consider $x,u,y\in\mathbb N_0$ and let $R(x,y)\Longleftrightarrow x<y$. Then we obtain $\exists x\forall u\exists y(x<x\vee x\geqslant u\vee(y<x\:\&\:y\geqslant u))$.
Let $x=n$ and let $u=n+1$. Our formula claims $\exists y(n<n\vee n\geqslant n+1\vee(y<n\:\&\:y\geqslant n+1))$. But that is obviously false: firstly, $n<n$ and $n\geqslant n+1$ both do not hold; secondly, the following system is inconsistent: $$\begin{equation*} \begin{cases} y<n\\ y\geqslant n+1 \end{cases} \end{equation*}$$
The formula is invalid, so it is unprovable.