My question is that whether the following statement is true or not.

In a complete metric space $(X, d)$, if a sequence of open balls $\{B(x_i, r_i)\}_{i=1}^\infty$ satisfies $$ \exists \epsilon > 0 ~~s.t.~ B(x_{i+1}, (1+\epsilon)r_{i+1}) \subset B(x_i, r_i), \forall i \ge 1 \tag{1} $$ then $\bigcap_{i=1}^\infty B(x_i, r_i) \neq \emptyset$.

Here are several related facts:

1. If we further require that $r_i \rightarrow 0$ as $i\rightarrow \infty$, then $\bigcap_{i=1}^\infty B(x_i, r_i) \neq \emptyset$ holds.

2. There is a example shows that $\bigcap_{i=1}^\infty B(x_i, r_i) = \emptyset$ if we replace $(1)$ by $\overline{B(x_{i+1}, r_{i+1})} \subset B(x_i, r_i)$ for all $i\ge 1$ where $\overline{B}$ is the closure of $B$

3. In a complete metric space, $B(x, r) \subset B(x', r')$ may hold for $r > r'$.

As Choi pointed it out, the above statement is similar to Cantor's intersection lemma. Recall that Cantor's intersection lemma in a complete metric space is given as follows:

In a complete metric space $(X, d)$, if a sequence of closed set $\{C_i\}_{i=1}^\infty$ satisfies $$ C_{i+1} \subset C_i, \forall i\ge 1 ~~\&~~ diameter(C_i) \rightarrow 0~ as~ i\rightarrow \infty $$ then $\bigcap_{i=1}^\infty C_i \neq \emptyset$

However, we don't require $diameter(B(x_i, r_i)) \rightarrow 0$ as $i\rightarrow \infty$. And there is indeed a example shows Cantor's inetersection lemma fails if we drop the diameter restriction. Please refer to Nested closed balls with empty intersection