Earlier I proved that a sequence in $\mathbb{R}^n$ converges to some limit $\vec{a}$ if and only if it is Cauchy. Now, I've also proved that if the sequence $\{\vec{x}_k\}$ is convergent to some limit $\vec{a} \in \mathbb{R}^n$, then $\lim\limits_{k\to\infty}\|\vec{x}_k - \vec{x}_{k+1} \| = 0$. I proved it in the following way: since $\{\vec{x}_k\}$ is convergent, it is Cauchy, which implies that, $\forall\varepsilon >0, \exists K\in \mathbb{N}$ such that $\|\vec{x}_k-\vec{x}_l\|<\varepsilon$ whenever $k,l\ge K$. Let $k>K,l=k+1$, then $k+1>K$ and $\|\vec{x}_k-\vec{x}_{k+1}\|<\varepsilon$, which implies that $\lim\limits_{k\to\infty}\|\vec{x}_k - \vec{x}_{k+1} \| = 0$ by definition of the limit at infinity.
Now, the next question I have to solve seems to contradict the above. Namely, the question asks me to provide a counterexample in $\mathbb{R}^n$ (for any $n\in\mathbb{N}$) to show that the converse of the statement above is not true. Because how can it be not true if Cauchy implies convergence?
"Because how can it be not true if Cauchy implies convergence?"
Because your counter-example will not/can not be Cauchy.
$\{a_n\}$ being Cauchy/convergent implies $|a_n - a_{n+1}|$ converges to $0$ but $|a_n - a_{n+1}|$ converging to $0$ does not imply $\{a_n\}$ is Cauchy/convergent.
And that is what you must find a counter example of: $\{a_n\}$ being such that $|a_n - a_{n+1}|$ converges to $0$ but $\{a_n\}$ is not Cauchy/convergent.
======= counter example below ======
Let $a_n = \sum_{i=1}^n \frac 1i$. Then $a_n$ does not converge but $|a_n - a_{n+1}| = \frac 1{n+1} \rightarrow 0$.
Cauchy is all $|a_n - a_m| \rightarrow 0$ for $n,m > M$ for some $M$. If only $|a_n - a_{n+1}|\rightarrow 0$ that is only the difference of adjacent terms. It does not imply Cauchy.
Obviously if all $n,m > M$ are such that $|a_n - a_m| < \epsilon$ then for all $n > M$ then $|a_n-a_m| < \epsilon$ and $n+1 > M$. But the converse isn't necessarily true. That something is true for all $n, n+1 > M$ in no way should it be true for all $n,m > M$.