squeeze theorem proof check 
let $\left \{ a_n \right \}$ , $\left \{ b_n \right \}$ , $\left \{ a_n \right \}$ be sequences of real numbers so that for all $n \in \mathbb{N}$ the inequalities $a_n \le b_n \leq c_n$ hold and so that $\lim_{x \rightarrow \infty } a_n = \lim_{x \rightarrow \infty } c_n$. Then $\left \{b_n \right \}$ converges and the limits are equal, that is, $\lim_{x \rightarrow \infty } a_n = \lim_{x \rightarrow \infty } c_n = \lim_{x \rightarrow \infty } b_n$.

My Proof:

Assume $\left \{ b_n \right \}$ does not converge. So irregardless of limit $L$, there exists $\epsilon>0$ for any $N\in \mathbb{N}$ so that some $n \ge N$ is such that $\left | b_n - L \right | \ge \epsilon$.
   So $-\epsilon \ge b_n - L \ge \epsilon$ and therefore $b_n \ge L + \epsilon$. Now suppose we have such an $\epsilon>0$ so that $b_n$ diverges. Pick an $N=max \left \{N_L,N_M \right \}$ where $N_L$ and $N_M$ are the required numbers such that for all $n\ge N_L$ or $n \ge N_M$, $\left | a_n -L \right | < \epsilon$ and $\left | c_n -M \right | < \epsilon$ (note that  $L=M$). This implies $c_n < L + \epsilon$. Hence $-c_n>-(L+\epsilon)$ , combine this with $b_n \ge L + \epsilon$ and we see that $b_n - c_n > 0$ so $b_n>c_n$ but this contradicts the initial assumption that $b_n\le c_n$ Therefore $b_n$ must be convergent.
  Because $b_n$ is convergent and for all $n\in \mathbb{N}$, $c_n\ge b_n$ it can be shown that $\lim_{n \rightarrow \infty } b_n \le \lim_{n \rightarrow \infty } c_n$. For similar reasons we also find that $\lim_{n \rightarrow \infty } a_n \le \lim_{n \rightarrow \infty } b_n$. Now because $\lim_{n \rightarrow \infty } a_n = \lim_{n \rightarrow \infty } c_n = L$ we have that $L \le \lim_{n \rightarrow \infty } c_n$ and $L \ge \lim_{n \rightarrow \infty } c_n$ which can only mean therefore that $\lim_{n \rightarrow \infty } b_n = L$

I am pretty confident in my proof and only used the "it can be shown" part because I wanted to mimize the amount of latex I had to write in order to post this.
So just looking for confirmation and if not then reasons why it may be wrong. Thanks in advance!
 A: In a schematic way, the squeeze theorem can be seen as
$$
\underbrace{
\overbrace{\big( \displaystyle\lim_{n\to\infty}a_n= L \big)}^{\mbox{true}} \mbox{ and }
\overbrace{\big( \displaystyle\lim_{n\to\infty}c_n= L \big)}^{\mbox{true}}  \mbox{ and }
\overbrace{\big((\forall n\in \mathbb{N}) \;\;a_n\leq b_n\leq c_n \big)}^{\mbox{true}}}_{\mbox{hypotheses}}
\implies 
\underbrace{\overbrace{\big( \displaystyle\lim_{n\to\infty}b_n= L \big)}^{\mbox{true}} }_{\mbox{thesis}}
$$
If I understand correctly you are using the Reductio ad absurdum technique. Knowing that the three hypotheses are true, suppose that the thesis is false and obtain as a consequence that some of the hypotheses is false (absurd). 
We recall the definition of $\lim_{n\to\infty}b_n= L$:
$$
\big( \displaystyle\lim_{n\to\infty}b_n= L \big)
:=
(\forall \epsilon >0)(\exists N_0\in \mathbb{N})
\big\lgroup
(\forall n\in\mathbb{N})(n>N_0)\implies (0<|b_n-L|<\epsilon)
\big\rgroup
$$
If $\big(\lim_{n\to\infty}b_n= L \big)$ is false, your negation $\big(\lim_{n\to\infty}b_n \neq L \big)$ is true. That is,
$$
(\exists \epsilon >0)(\forall N_0\in \mathbb{N})
\big\lgroup
(\exists n\in\mathbb{N})(n>N_0)\mbox{ and }  
\big\lgroup
(|b_n-L|\geq\epsilon)\mbox{ or }
(|b_n-L|\leq 0)
\big\rgroup
\big\rgroup.
$$
In your case, in the notation and symbology of this answer, you did the logical negation as follows
$$
(\exists \epsilon >0)(\forall N_0\in \mathbb{N})
\big\lgroup
(\exists n\in\mathbb{N})(n>N_0)\mbox{ and }  (-\epsilon \ge b_n - L \ge \epsilon\big\rgroup.
$$
At least that's what I understood in

"So $-\epsilon \ge b_n - L \ge \epsilon$ and therefore $b_n \ge L + \epsilon$.'' 

This is simply not the correct way to make the logical negation of a proposition.
A: As indicated by MathOverview, your proof needs some improvement. You should state clearly (and you came close) that if $b_n$ fails to converge to $L$, then there exists some $\epsilon$ such that $b_n > L + \epsilon$ or $b_n < L - \epsilon$ for infinitely many $n$. However, for all sufficiently large $n$ we have $$L - \epsilon < a_n \leqslant c_n < L + \epsilon$$  If $b_n > L + \epsilon$ we get the contradiction $b_n > c_n$, as you showed.  You then need to show that $b_n < L - \epsilon$ leads to the contradiction $b_n < a_n$ in a similar way.
It would be easier to argue directly that 
$$-\epsilon < a_n -L < \epsilon \,\,\,
\,\,\,\,\,\,\text{if} \,\,\,n \geqslant N_a \\ -\epsilon < c_n -L < \epsilon \,\,\,\,\,\,\,\,\text{if} \,\,\,n \geqslant N_c $$ 
Consequently, if $n \geqslant \max(N_a,N_c),$ then 
$$-\epsilon < a_n-L \leqslant b_n-L \leqslant c_n - L < \epsilon \\ \implies |b_n - L| < \epsilon$$
