I think this statement is true.

Suppose we had a counterexample $\{B(x_i,r_i)\}_{i=1}^\infty$ satisfying condition (1) for some $\epsilon > 0$ but whose intersection was empty. Observe that any subsequence will still be a counterexample.

For each $i$, the point $x_i$ does not belong to some $B(x_j,r_j)$, as otherwise the intersection of all the balls would be nonempty. So by passing to a subsequence we can ensure that $x_i \not\in B(x_{i+1},r_{i+1})$ for all $i$. Note that this forces $r_{i+1} < r_i$, as $x_{i+1}$ does belong to $B(x_i,r_i)$.

By Cantor's intersection lemma, the $r_i$ cannot decrease to zero. So they must decrease to some $r > 0$. Without loss of generality we can now assume that $(1 + \epsilon)r > r_1$. But for any $i > j$ we have $x_i \in B(x_j,r_j)$, i.e., $d(x_i,x_j) < r_j \leq r_1 < (1+\epsilon)r$, which means that every $x_i$ belongs to every $B(x_j,(1+\epsilon)r_j)$. But then by condition (1) every $x_i$ belongs to every $B(x_j,r_j)$, a contradiction.

I think this statement is true.

Suppose we had a counterexample $\{B(x_i,r_i)\}_{i=1}^\infty$ satisfying condition (1) for some $\epsilon > 0$ but whose intersection was empty. Observe that any subsequence will still be a counterexample.

For each $i$, the point $x_i$ does not belong to some $B(x_j,r_j)$, as otherwise the intersection of all the balls would be nonempty. So by passing to a subsequence we can ensure that $x_i \not\in B(x_{i+1},r_{i+1})$ for all $i$. Note that this forces $r_{i+1} < r_i$, as $x_{i+1}$ does belong to $B(x_i,r_i)$.

By Cantor's intersection lemma, the $r_i$ cannot decrease to zero. So they must decrease to some $r > 0$. Without loss of generality we can now assume that $(1 + \epsilon)r > r_1$. But for any $i$ we have $x_i \in B(x_1,r_1)$, i.e., $d(x_i,x_1) < r_1 < (1+\epsilon)r$, which means that $x_1$ belongs to every $B(x_i,(1+\epsilon)r_i)$. But then by condition (1) $x_1$ belongs to every $B(x_i,r_i)$, a contradiction.