How to prove the Squeeze Theorem for sequences
Let $\varepsilon > 0$. Since $x_n \to l$, there exists $N_1 = N_1(\varepsilon)$ such that $|x_n - l| < \varepsilon$ for all $n \ge N_1$. Since $z_n \to l$, there exists $N_2 = N_2(\varepsilon)$ such that $|z_n - l| < \varepsilon$ for all $n \ge N_2$. Set $N = \max\{N_1,N_2\}$. If $n \ge N$, then $$y_n - l \le z_n - l < \varepsilon$$ and $$y_n - l \ge x_n - l > -\varepsilon$$ Hence $|y_n - l| < \varepsilon$ for all $n \ge N$. Since $\varepsilon$ was arbitrary, $y_n \to l$.
Write
$$|y_n-l|\le |y_n-x_n|+|x_n-l|\le( z_n-x_n)+|x_n-l|$$ Can you take it from here?
Ok, so this is what I have now:
Let $x_n \le y_n \le z_n \space\space \forall n \in \mathbb N$ and $\lim_{n \to \infty} x_n = \lim_{n \to \infty} z_n = l$ for some $l \in \mathbb R$.
Then we have: $$\begin{align} &\forall \varepsilon_1 > 0,\, \exists N_{\varepsilon_1} \in \mathbb N,\, \forall n \ge N_{\varepsilon_1}\space |x_n - l| < \varepsilon_1 \\ &\forall \varepsilon_2 > 0,\, \exists N_{\varepsilon_2} \in \mathbb N,\, \forall n \ge N_{\varepsilon_2}\space |z_n -l| < \varepsilon_2 \end{align}$$ Let $N = \max\lbrace N_{\varepsilon_1},N_{\varepsilon_2} \rbrace$ and $\varepsilon = \min\lbrace \varepsilon_1, \varepsilon_2 \rbrace$, so that we have $$\forall \varepsilon > 0,\, \exists N \in \mathbb N,\, \forall n \ge N\space |x_n - l| < \varepsilon \space \text{and} \space |z_n -l| < \varepsilon$$
Let $\frac\varepsilon3 > 0$. Then $\exists N \in \mathbb N,\, \forall n \ge N\space |x_n -l| < \frac\varepsilon3 \text{ and } |z_n -l| < \frac\varepsilon3$ so that $$|z_n - x_n| = |z_n - l + l - x_n| \le |z_n - l| + |x_n - l| < \frac\varepsilon3 + \frac\varepsilon3 = \frac{2\varepsilon}3$$ By assumption, $$\begin{align} x_n \le&\, y_n \le z_n \\ 0 \le&\, y_n - x_n \le z_n - x_n \\ &|y_n - x_n| \le |z_n - x_n| \\ |y_n - l| = |y_n - x_n + x_n - l| \le& |y_n - x_n| + |x_n - l| \le |z_n - x_n| + |x_n - l| < \frac{2\varepsilon}3 + \frac\varepsilon3 = \varepsilon \end{align} $$ And so $\lbrace y_n \rbrace$ also converges to $l$, thus, Q.E.D.
I think that that's satisfactory. If there are any ambiguities or mistakes, please let me know. Thanks to everybody for your advice and contributions.