Show a sequence such that $\lim_{\ N \to \infty} \sum_{n=1}^{N} \lvert a_n-a_{n+1}\rvert< \infty$, is Cauchy 
Attempt. Rewriting this we have, $$\sum_{n=1}^{\infty} \lvert a_n-a_{n+1}\rvert<  \infty
\,\,\,\Longrightarrow\,\,\, \exists N \in \mathbb{N}\ \ s.t,\  \ \sum_{n \geq N}^{\infty} \lvert a_n-a_{n+1}\rvert < \infty$$
Taking $m>n+1$ we have, $$\sum_{m > n \geq N}^{\infty}\lvert a_n-a_m\rvert< \infty$$
From the above statement we have: $\,\lim_{\ n \to \infty} \lvert a_n-a_m\rvert=0$.
Now combining out statement we have, $\forall \varepsilon>0, \exists N \in \mathbb{N}\ s.t \ \lvert a_n-a_m\rvert< \varepsilon, \forall, m\geq n \geq N$.

The third step is not that clear to me. However, you could do the following:
Setting $U_N\stackrel{\rm def}{=}\sum_{n=1}^N \lvert a_n - a_{n+1}\rvert$, $(U_N)_N$ is real-valued and convergent, and therefore Cauchy. Let $\varepsilon > 0$ be arbitrary: there exists $N$ such that, for all $m > N$ and $p\geq 0$,
$$
\lvert U_m - U_{m+p} \rvert \leq \varepsilon
$$
Rewriting, this means $\sum_{n=m}^{m+p} \lvert a_n - a_{n+1}\rvert \leq \varepsilon$. By the triangle inequality,
$$
\lvert a_m - a_{m+p} \rvert = \lvert \sum_{n=m}^{m+p} a_n - a_{n+1}\rvert \leq \sum_{n=m}^{m+p}\lvert a_n - a_{n+1}\rvert \leq \varepsilon
$$
for all $m > N$ and $p\geq 0$.
