I want to prove that $\forall x \in \mathbb{R}\, \exists M \in \mathbb{Z}\, M \le x < M + 1.$
We know that $\forall x \in \mathbb{Q}\, \exists M \in \mathbb{Z}\, M \le x < M + 1.$
We also know that $\forall q_1,q_2 \in \mbox{fundamental}(\mathbb{Q}^{\mathbb{N}})\, (\exists N \in \mathbb{N}\, \forall n > N\, q_1(n) \le q_2(n)) \implies [q_1]_{\mathbb{R}} \le [q_2]_{\mathbb{R}}\, (*)$
Given $x \in \mathbb{R}$ $x = [x_n]_{~\mathbb{R}}$ where $x_{n}\colon \mathbb{N} \to \mathbb{Q}$ and $x_{n}$ is fundamental in ${\mathbb{Q}}.$ that is, $\forall \varepsilon > 0\, \exists N \in \mathbb{N}\, \forall m,n \in \mathbb{N}_{>N}\, |x_{m}-x_{n}| < \varepsilon.$
Since its fundamental it follows that $\forall n > N_{0}\, |x_{n_0}| - 1 < |x_{n}| < |x_{n_0}| + 1$.
Since, $|x_{n_0}| \in \mathbb{Q}$ it follows that there exists $M \in \mathbb{Z}$ such that $M \le |x_{n_0}| < M + 1$.
Combining two of the above inequalities we get
$$\forall n > N_{0}\, M - 1 < |x_{n}| < M + 2$$ thus from $(*)$ $$M - 1 \le x \le M + 2$$
From this and the trichotomy of order used on $[x,M-1,M,M+1,M+2]$ we eventually get $$\exists M \in \mathbb{Z}\, M \le x < M + 1.$$