Claim: If $(x_n)_{n=0}^\infty$ a Cauchy sequence, then for any $N > 0$ we have $(x_n)_{n=N}^\infty$ is Cauchy as well.
Proof: Fix $\epsilon > 0$. Since $(x_n)_{n=0}^\infty$ Cauchy, there exists $M$ so that for $n,m \geq M$ we have $d(x_n, x_m) < \epsilon$. If $M \geq N$, then choose the same $M$ for $(x_n)_{n=N}^\infty$. If $N > M$, then notice that for all $n,m \geq N > M$ we have that $d(x_n, x_m) < \epsilon$, so there still exists such an $N$. So if we let $N' = \max\{N, M\}$, then we have that for all $n,m \geq N'$ $d(x_n, x_m) < \epsilon$, proving the subsequence $(x_n)_{n=N}^\infty$ is Cauchy.
Claim: Suppose that for every sequence of nested closed balls $\overline{B}(x_{n+1}, r_{n+1}) \subseteq \overline{B}(x_n, r_n)$ with $r_n \rightarrow 0$, we have $\bigcap_{n=1}^\infty \overline{B}(x_n, r_n) \neq \varnothing$. If $(x_n)$ is Cauchy, then there is an $x$ so that $x_n \rightarrow x$.
Proof:
Take $(x_n)$ a Cauchy sequence. Notice there exists an $N_1$ so that for all $n,m \geq N_1$, we have $d(x_n, x_m) < 1/2$. Let $z_1 = x_{N_1}$. Notice that $(x_n)_{n=N_1}^\infty \subseteq \overline{B}(z_1, 1)$ is a Cauchy sequence by the above claim. Thus there exists an $N_2 > N_1$ so that for all $n,m \geq N_2$, we have $d(x_n, x_m) < 1/4$. Let $z_2 = x_{N_2}$. We claim that $\overline{B}(z_2, 1/2) \subseteq \overline{B}(z_1, 1)$. Taking $y \in \overline{B}(z_2, 1/2)$, we see
$$d(y,z_1) \leq d(y,z_2) + d(z_2, z_1) \leq 1/2 + 1/2 = 1,$$
so $y \in \overline{B}(z_1, 1)$. Continue inductively like this. Suppose we have $z_r$ and $N_r$ so that $(x_n)_{n=N_r}^\infty \subseteq \overline{B}(z_r, 1/2^r)$ and for all $n,m \geq N_r$ we have $d(x_n, x_m) < 1/2^{r+1}$. This sequence is Cauchy, and so we find $N_{r+1} > N_r$ so that for $n,m \geq N_{r+1}$ we have $d(x_n, x_m) < 1/2^{r+2}$. Letting $z_{r+1} = x_{N_{r+1}},$ we take $\overline{B}(z_{r+1}, 1/2^{r+1})$. We claim that $\overline{B}(z_{r+1}, 1/2^{r+1}) \subseteq \overline{B}(z_r, 1/2^r)$. Take $y \in \overline{B}(z_{r+1}, 1/2^{r+1})$, and note
$$d(z_r, y) \leq d(z_r, z_{r+1}) + d(z_{r+1}, y) \leq \frac{1}{2^{r+1}} + \frac{1}{2^{r+1}} = \frac{1}{2^r},$$
so $y \in \overline{B}(z_r, 1/2^r)$.
Using this, we have constructed a decreasing sequence of closed balls, so using the property we get that $\bigcap_{r=0}^\infty \overline{B}(z_r, 1/2^r) \neq \varnothing$. Take $x$ in the intersection. We claim $x_n \rightarrow x$. This will follow if we show that the subsequence $z_r \rightarrow x$. To see this, fix $\epsilon > 0$. There exists $R$ sufficiently large so that $1/2^R < \epsilon$, so for $r \geq R$ we have that $d(z_r, x) \leq 1/2^r < \epsilon$ (since $x \in \overline{B}(z_r, 1/2^r)$). Thus $z_r \rightarrow x$, and since we have a convergent subsequence of a Cauchy sequence this implies that $x_n \rightarrow x$.
For this last fact, see Cauchy sequence is convergent iff it has a convergent subsequence.
