Is my attempt at proving this correct ?
Proof. For the sake of contradiction, suppose that there exists $x\in\mathbb{R}$ such that $x\geq n\in\mathbb{N}$. Since $\mathbb{N}$ is always increasing, it follows by MCT that $\mathbb{N}$ converges to some limit $L$. That is, for any positive $\epsilon\in\mathbb{R}$, there exists $N\in\mathbb{N}$ such that for all $n\in\mathbb{N} > N$, $$|n-L| < \epsilon$$ Since $\mathbb{N}$ is always increasing, if $L$ is less than any element $n$ in $\mathbb{N}$, it will be less than all the infintely many elements greater than $n$, but then it would not be the limit to which $\mathbb{N}$ converges. Therefore, it must be the case that $L\geq n\in \mathbb{N}$ and so $|n-L| = L-n$. Hence, we now we have the following inequality $$L-n < \epsilon$$ Choose $\epsilon =1$ $$L < n+1$$ Since $(n+1)\in\mathbb{N}$, this is a contradiction. Thus, our assumption that there exists $x\in\mathbb{R}$ such that $x\geq n\in\mathbb{N}$ is false.