Problem: If sequence $ (a_n) $ has 10,-10 as partial limits and in addition $ \forall n \in \mathbb{N}.|a_{n+1} - a_n| \leq \frac{1}{n} $, then 0 is a partial limit of $ (a_n) $.
Answer (Proof): We need to show $ \forall \epsilon > 0 \forall N \in \mathbb{N} \exists n \geq N . | a_n| < \epsilon $.
Let $ \epsilon >0 $ and $ N \in \mathbb{N}$ be arbitrary. We know $ \frac{1}{n} \rightarrow 0 $, Hence there exists $ N_1 \in \mathbb{N}$ s.t. $ \forall n>N_1 . \frac{1}{n} < \epsilon $ . Since -10 is a partial limit then there exists $ n_1 \geq max \{ N, N_1 \} $ s.t. $ -1 < a_{n_1} - ( -10 ) < 1 $ , specifically $ a_{n_1} < -9 $ . Since 10 is also a partial limit, there exists $ n_2 > n_1 $ s.t. $ -1 < a_{n_2} - 10 < 1 $, specifically $ a_{n_2} > 9 $. Since $ a_{n_1} < 0 $ and $ a_{n_2} > 0 $ and also $ n_2 > n_1 $, $ \color{red}{ \text{there exists} } $ $ n_1 \leq n < n_2 $ $ \color{red}{ \text{s.t.} } $ $ a_n \leq 0 $ , $ a_{n+1} > 0 $.
In addition $ n \leq N_1 $, hence $ a_{n+1} < a_{n+1} - a_{n} = | a_{n+1} - a_{n} | \leq \frac{1}{n} < \epsilon $ and $ n+1 \geq N $, as we wanted.
My Difficulty: I understood everything up to the line "$ \color{red}{ \text{there exists} } $ $ n_1 \leq n < n_2 $ $ \color{red}{ \text{s.t.} } $ $ a_n \leq 0 $ , $ a_{n+1} >0 $. "
I don't understand why such an $ n $ exists s.t. it satisfies $ a_n \leq 0 $ , $ a_{n+1} >0 $. Do we get this $ n $ from some kind of instantiation here? ( Maybe from instantiation in the definition for the partial limits 10 & -10 ? )
Related: Prove $0$ is a partial limit of $a_n$ - In this linked thread first answer is virtually the same proof above here but my question remains the same because I have still don't understand from there why such an $ n $ exists.