[Math] Does this prove the Squeeze Theorem

I'm aware that there are plenty of proofs for the squeeze theorem but I wanted to verify if I'm on the right track for this approach.

Problem: Show that if $x_n \leq y_n \leq z_n \hspace{1mm} \forall n \in \mathbf{N}$ and if lim$(x_n)$ = lim$(z_n) = \ell$, then lim$(y_n) = \ell$ as well.

Solution: Given that $x_n \leq y_n \leq z_n \hspace{1mm} \forall n \in \mathbf{N}$ then $y_n$ converges otherwise $x_n$ or $z_n$ diverge. Let lim$(y_n) = y$. Then,

\begin{align} &x_n \leq y_n \leq z_n \\ &0 \leq y_n-x_n \leq z_n-x_n \\ \implies &\text{lim}(y_n-x_n) = y-\ell \leq \text{lim}(z_n-x_n) = \ell-
\ell = 0 \\ \implies &y \leq \ell
\end{align}

Similarly,
\begin{align} &x_n-z_n \leq y_n-z_n\leq 0 \\
\implies&\text{lim}(x_n-z_n) = \ell-\ell = 0 \leq \text{lim}(y_n-z_n)=y-\ell \\ \implies&\ell \leq y
\end{align}

Hence, $y=\ell \hspace{1cm}\square$

Okay so I used @Fred's hint and changed the proof.

$\epsilon < x_n-\ell \leq y_n-\ell \leq z_n -\ell < \epsilon$

since lim$(x_n)$ = lim$(z_n)= \ell$

Thus $-\epsilon < y_n – \ell < \epsilon \implies |y_n-\ell|<\epsilon$

[Math] Does this prove the Squeeze Theorem

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