Assume that $\{x_n\}$ is a divergent sequence and $L$ is a real number.
Prove that there exists $ \epsilon \gt 0$ and a subsequence of $\{x_n\}$ like $\{x_{p_n}\}$ such that for each $n \in \mathbb N$, $|x_{p_n}-L| \gt \epsilon$ .
My try :
I assumed that its not true. So every subsequence of $\{x_n\}$ has this property :
For every positive $\epsilon$, In every subsequence, There exists an element like $E$ such that $|E-L|\le \epsilon$.
So, If i take $E_i$'s and make a new subsequence, that new subsequence converges to $L$. I don't know why but i think it should lead to a contradiction.
I assume your definition of "divergent" is "non-convergent". The direct negation of the statement you want to prove is: There exists a number $L$ such that for all $\epsilon>0$ and for all subsequences $(x_p)_{p\in \mathbb N}$ there exists an $n$ such that $|x_n-L|<\epsilon$. Now you reason as follows. Start with any subsequence $(x_p)$. There is an $x_{p_1}$ having distance less than $1$ to $L$. Take that $x_{p_1}$ out to obtain a new subsequence. There is a $x_{p_2}$ having distance $\frac 12$ to $L$ take it out and repeat the reasoning (for 1/3, ...). Then the sequence $x_{p_i}$ is a subsequence of $(x_p)$ having a unique limit point $L$. This tells you that every subsequence has $L$ as a limit point. But this means that the biggest and smallest limit point (which exist!) of your original sequence equal $L$, hence it is convergent. Contradiction!