Let
$$X:=\left\{(x_n)_{n \in \mathbb{N}}: x_n \in \mathbb{R},\sum_{n=1}^\infty \frac{1}{2^n} \cdot \frac{|x_n|}{1+|x_n|} \lt \infty\right\}$$
be the metric space of sequences with the metric:
$$d(x,y) := \sum_{n=1}^\infty \frac{1}{2^n} \cdot \frac{|x_n-y_n|}{1+|x_n-y_n|}$$
Prove that the unit ball entered around 0, $B_1(0):= \left\{(x_n)_{n \in \mathbb{N}}:\sum_{n=1}^\infty \frac{1}{2^n} \cdot \frac{|x_n|}{1+|x_n|} \leq 1 \right\}$ is sequentially compact, and hence compact.
I don't get how it can be sequentially compact, for example the sequences $(1,1,1,\ldots), (2,2,2,\ldots) \ldots, (k,k,k,\ldots) \ldots,\forall k \in \mathbb{Z}^{+}$ lie inside the ball the minimum distance between any two sequences in the set is $\frac{1}{4}$ ? Did I think about it wrongly...
Looks fine to me, except I'd say that the minimum distance is $\frac{1}{2}$, not $\frac{1}{4}$. Note that \begin{align*} d((a, a, a, \ldots), (b, b, b, \ldots)) &= \sum_{n=1}^\infty \frac{1}{2^n}\frac{|a - b|}{1 + |a - b|} \\ &= \frac{|a - b|}{1 + |a - b|}\sum_{n=1}^\infty \frac{1}{2^n} \\ &= \frac{|a - b|}{1 + |a - b|} \\ &\ge \frac{1}{1 + 1} = \frac{1}{2}, \end{align*} with equality achieved when $|a - b| = 1$.
I agree with the conclusion. This is not sequentially compact. Your sequence is definitely in the unit ball around the $0$ sequence, and this shows that no subsequence can be Cauchy, let alone convergent.