Is my proof that $\lim\limits_{n\to +\infty}\frac{u_{n+1}}{u_n}=1$ correct?

Not really an answer to your question, but an alternative approach.

If $u_n\to \ell_1$ and $v_n\to \ell_2$ with $l_1\neq 0$, then $\frac{v_n}{u_n}\to \frac{\ell_2}{\ell_1}$ as a general rule. So if $v_n=u_{n+1}$...

As an exercise, we give a detailed argument directly from the definition. Suppose that the sequence $(u_n)$ has limit $a\gt 0$. We want to show that for any $\epsilon\gt 0$, there is an $N$ such that $$1-\epsilon\lt \frac{u_{n+1}}{u_n}\le 1\tag{1}$$ if $n\gt N$. Note that $$\frac{u_{n+1}}{u_n}\ge \frac{a}{u_n},$$ so it suffices to make $\frac{a}{u_n}\gt 1-\epsilon$. This will be the case automatically if $\epsilon\ge 1$, so we may suppose that $\epsilon\lt 1$.

For $0\lt \epsilon\lt 1$ we have $$\frac{a}{u_n}\gt 1-\epsilon \quad\text{iff}\quad u_n \lt \frac{a}{1-\epsilon} \quad\text{iff}\quad u_n-a\lt \frac{a\epsilon}{1-\epsilon}.$$

Since the sequence $(u_n)$ converges to $a$, there is an $N$ such that if $n\gt N$, then $u_n-a\lt \frac{a\epsilon}{1-\epsilon}$. For any such $n$, Inequality (1) will hold.

Remark: Informally, this is simpler than it looks. We can scale the sequence $(u_n)$ so that it has limit $1$. That does not change ratios.