Proof of Part 1: Let $(x_n)$ be a sequence that converges to x. In other words, for all $n \ge N \in \mathbb{N}$, $d(x_n,x) < \epsilon$, for all $\epsilon > 0$. We choose $\epsilon$ as small as we like, so we pick $\epsilon \over 2$. $\therefore$ , $\forall\ m \ge N$ $d(x_m,x)\le \frac{\epsilon}{2}$. By Triangle inequality: $d(x_n, x_m) \le d(x_n, x) + d(x, x_m) \implies d(x_n, x_m) < \frac {\epsilon}{2} + \frac {\epsilon}{2} = \epsilon$.
Proof of Part 2: Let $(x_n)$ be a Cauchy sequence. So for some $N \in \mathbb{N}$, $(x_n: n \ge N) \subset B_\epsilon (x_N)$, for some $\epsilon > 0$. Consider instead $B_{r+\epsilon} (x_N)$, where $r = \max_{1\le i \le N} \ (d(x_N, x_i))$. Then all elements of $(x_n)$ are within $B_r (x_N)$.
I feel confident in the idea I have of the proof (drew out the circles of the balls and what not) but would appreciate feedback on the writing and if the idea is sound.