I have two doubts on the initial statements on this proof of the completeness of $l^1$.
I don't understand why we can say that the sequence $x_k^{(n)}$ is a Cauchy sequence in $R$. I understand that $|x_k^{(n)} - x_k^{(m)}| \le \epsilon$, my doubt is that $x^{(n)}$ and $x^{(m)}$ are two different sequences.
When defining a Cauchy sequence in $R$ we say that a sequence is Cauchy if if for every positive real number $\epsilon_1$, there is a positive integer $N$ such that for all natural numbers $m, n > N$ $$|x_m - x_n| < \varepsilon $$
In this case the distance is between elements of the same sequence. In the proof we have that the components of different sequences can be made arbitrary small. I hope I have made clear what is confusing me.
The other doubt is that I do not understand why we suspect $X$ to be the limit of the sequence $X_n$, what makes us suspect that? Maybe solving my previous doubt will help me understand this.
I'm going to start from the general, then move to the specific, because part of the confusion is the notation of sequences of sequences.
In general, let $U,V$ be two metric spaces. Let $f:U\to V$ be defined so that $$d_V(f(u_1),f(u_2))\leq d_U(u_1,u_2)$$ for all $u_1,u_2\in U$. It is easy to prove that such a function is continuous.
Under these conditions, if $\{u_i\}$ is a Cauchy sequence in $U$, then we can show that $\{f(u_i)\}$ is a Cauchy sequence in $V$.
Now, in the above case, $U=\ell^1$ and $V=\mathbb R$, and $f=f_k$ is defined on $\ell^1$ as:
$$f_k(\{x_i\}_{i=1}^\infty) = x_k$$
All that the proof above is saying is that $f_k$ has this property, and thus if $X_1,\dots,X_n,\dots$ is Cauchy in $\ell^1$ then $f_k(X_1),\dots,f_k(X_n),\dots$ is Cauchy in the real numbers.
As for the question about why we "suspect." Go back to the general case. If $V$ is complete, and $u_1,\dots,u_n,\dots$ is Cauchy in $U$, then if it converges to $u\in U$, we'd have to have $\lim f(u_n)=f(u)$ since such $f$ is continuous.
In our specific case, if $X_n$ converges to $X\in\ell^1$, we'd have to have $f_k(X_n)\to f_k(X)$ for all $k$ - that is, we'd have to get the same result as component-wise convergence.
That's actually stronger than suspicion, then, because we see that the "component-wise" $X$ is the only possible candidate for the limit. But it still doesn't prove that it is the limit. We have not even yet shown that the component-wise limit is in $\ell^1$.