Set of subsequential limit - Trying to finish a proof I'm trying to answer this proof:
set of subsequential limit
I followed the int that was given, so I made  $\varepsilon := \dfrac{\min\{|c-a|, |c-b|\}}{2}$
Then, individually I tried to solve $ c - \varepsilon < x_n <c + \varepsilon $ making $\varepsilon := \dfrac{{|c-a|}}{2}$ and $\varepsilon := \dfrac{{|c-b|}}{2}$
And I reached that $ |b|< c $ and $ |a|< c $ 
But now how can I conclude what I want to prove... Can someone give a hint? Or am I proving this wrongly?
Thanks! 
 A: The idea behind the proof that you have linked to is that you choose a value $c \neq a, b$ and show that it cannot be a subsequential limit. 
We have $\displaystyle\lim\limits_{k \rightarrow \infty}{x_{2k}} = a$, implies for any $\epsilon > 0$ there exists $N_{a} \in \mathbb{N}$ such that for for all $n \geq N_{a}$, $|x_{2n} - a| < \epsilon$
Similarly by $\displaystyle\lim\limits_{k \rightarrow \infty}{x_{2k + 1}} = b$, for any $\epsilon > 0$ we have an $N_{b} \in \mathbb{N}$ such that for all $n \geq N_{b}$, $|x_{2n + 1} - b| < \epsilon$
The point of the hint you refer to is that, if we take $\epsilon \leq \displaystyle\min\left\{\cfrac{|a - c|}{2}, \cfrac{|b - c|}{2}\right\}$ and let $N = \displaystyle\max\left\{N_{a}, N_{b}\right\}$ then both subsequences, $\left(x_{2k}\right)_{k}$ and $\left(x_{2k + 1}\right)_{k}$ remain in the union of intervals $(a - \epsilon, a + \epsilon) \cup (b - \epsilon, b + \epsilon)$ for all $n \geq N$. 
Then, because of how we chose $\epsilon$, $c$ is not in this union of intervals and therefore cannot be in the set of subsequential limits. 
