Given a cauchy-sequence $\{x_i\}_{i\in \mathbb{N}}$ in a normed space $X.$ I need to construct a series that converges in $\mathbb{R}$ with $\{y_i\}_{i\in \mathbb{N}}$ a sequence in $X$:
$$\sum_{i=1}^\infty \lVert y_i\rVert$$
The most natural (for me) series to look at is the one where $y_i:=x_i-x_{i-1}$. Because then the cauchy-ness of $\{x_i\}_{i\in \mathbb{N}}$ can be directly translated to a condition on $y_i$:
$$\lim_{i \rightarrow \infty} y_i =0$$
I loose some information here because I only use that $\lVert x_{i}-x_{i-1} \rVert$ goes to zero, but the same holds if we do not take consecutive indices, which gives that
$$\lim_{n \rightarrow \infty}\lVert\sum_{i=n}^{n+k} y_i \rVert=0$$
for any fixed $k \in \mathbb{N}$.
How could I use these properties to get that for any $\epsilon>0$ there is a $N$ such that for all $n,m>N$: $\sum_{i=n}^m \lVert y_i\rVert<\epsilon$? The lack of upperbound on $m$ gives me problems since we get infinitely many terms. Also the fact that $\lVert\sum_{i=n}^{m} y_i \rVert \leq \sum_{i=n}^m \lVert y_i\rVert$ confuses me as my reflex to try the triangle inequality leads me nowhere.
You're on the right track, but some adjustments are required. If you're not sure why the argument cannot work for you in its current state, consider the construction for the counterexample provided in the accepted answer here.
(Basically, in any non-complete normed space, pick a sequence 'converging' to 'a hole'. If you define $(y_n)$ in the aforementioned manner, then the $x_n$'s are the partial sums of $\sum y_n$, which is to say the series cannot converge. However, the convergence of the series of norms $\sum\|y_n\|$ depends solely on the rate at which $x_n$ tends to the 'hole', so you could easily generate a counterexample from any such sequence.)
Of course, there are also many examples of $\sum y_n$ converging where $\sum\|y_n\|$ diverges (such as for $x_n = \frac{(-1)^n}{n}\in\mathbb{R}$).
Hint: So, to conclude the proof the way you started, try to first deal with Cauchy sequences where the distance 'decays' very fast--say exponentially--so that you'd be able to show $\sum\|y_n\|<\infty$. Only then, generalize the result to any Cauchy sequence.
Added More than enough time has passed, so in the interest of completeness (no pun intended) I'll conclude the proof:
One takes an arbitrary Cauchy sequence $(x_n)\subset X$. For all $m\in\mathbb{N}$ there exists $n_m$ such that $l,k\geq n_m\implies |x_l-x_k|<2^{-m}$, and w.l.o.g. we can pick an increasing sequence $(n_m)$. We can therefore let $y_m:=x_{n_{m+1}}-x_{n_m}$. Note: $$\sum_{m\geq 0}\|y_m\|\leq \sum_{m\geq 0}2^{-m} = 1,$$ implying that the sequence $(x_{n_{N+1}}-x_{n_0}) = \sum_{m=0}^N y_m$ converges in $X$. Hence $(x_{n_N})_{N\in\mathbb{N}}$ converges, and we're left to verify that a Cauchy sequence having a convergent subsequence converges to the same limit.