Given: $(a_n), (b_n)$, and $(c_n)$ are sequences, with $a_n \le b_n \le c_n$ for all n. Also, $a_n \to a$ and $c_n \to a$.
Prove by contradiction: $(b_n)$ converges and $b_n \to a$.
Here is my attempt. Please let me know if this is a viable proof, and how I can improve upon it.
Proof:
Suppose $(b_n)$ does not converge to $a$.
Then, $\exists \epsilon _b \gt 0$ $ \forall M \in \mathbb N$ such that $\forall n \ge M, |b_n-a| \ge \epsilon _b$
and $\forall \epsilon \gt 0, \exists N_1, N_2 \in \mathbb N$ such that $\forall n \ge N_1, |c_n-a| \lt \epsilon$ and $\forall n \ge N_2, |a_n-a| \lt \epsilon$.
Let $N=max(N_1,N_2)$.
Letting $M=N$, we get:
$\exists \epsilon _b \gt 0$ such that $\forall n \ge N$, $|b_n-a| \ge \epsilon _b, |c_n-a| \lt \epsilon _b, |a_n-a| \lt \epsilon _b$.
Thus:
$a-\epsilon _b \lt a_n $
$a-\epsilon _b \gt c_n$
$a-\epsilon _b \ge b_n$ or $a+\epsilon _b \le b_n$
As a result, $\exists \epsilon _b \gt 0$ such that $\forall n \ge N, b_n \le a-\epsilon _b \lt a_n$ or $c_n \lt a + \epsilon _b \le b_n$, contradicting the fact that $a_n \le b_n \le c_n$. Therefore, $b_n$ must converge to $a$.
Best Answer
I think that your initiating implication is invalid:
A non-converging sequence $b_n$ might well have infinite points arbitrarily close to $a$ (without converging to it).
Proof: Since $a_n \leq b_n \leq c_n$ then $0\leq b_n-a_n\leq c_n-a_n$, thus $|b_n-a_n|\leq c_n-a_n$.
Combining the above with the fact that $\lim(c_n-a_n)=a-a=0$ we get: $\lim(b_n-a_n)=0$.
Now we can write the terms of $(b_n)$ as the sum of the terms of two converging sequences: $b_n=(b_n-a_n)+a_n$, so we have: $$ \lim b_n=\lim\big((b_n-a_n)+a_n \big)=\lim(b_n-a_n)+\lim a_n=0+a=a $$