So here is my understanding. We know that every rational Cauchy sequence is bounded by some $N \in \mathbb{N}$ such that for all k, $|x_k|\leq N$. This is true because there exists m such that $|x_j-x_k|\leq1$ for $j,k\geq m$. We need to choose $N$ larger than $|x_m| + 1$.
So I think we could have that $x_k\leq N < x$ where $x$ is a real number.
But I am struggling to understand the relationship between $N$ and $x$. $N$ is just an upper bound based on the rational Cauchy sequence. $x$ is the equivalence class of the rational Cauchy sequence.
If a sequence converges in $\Bbb R$, then it is a Cauchy sequence.
So, for a given real $x$, you only have to show up a sequence $x_n$ of rationals, such that $x_n$ converges to $x$ from below.
Hint: You can choose $x_0, x_1,x_2,\dots \in\Bbb Q$ such that $0<x-x_n<\dfrac1{10^n}$.
Note that we are in the middle of constructing $\Bbb R$ from $\Bbb Q$ by Cauchy competition.
In general, a Cauchy sequence in a metric space is a sequence that 'wants to converge' to some point (which might not yet be present in the given space), and you can think of the Cauchy completion as simply filling in those 'holes'.