How to prove the Squeeze Theorem for sequences

The formulation I'm looking at goes: If $\lbrace x_n\rbrace$, $\lbrace y_n\rbrace$ and $\lbrace z_n \rbrace$ are sequences such that $x_n \le y_n \le z_n$ for all $n \in \mathbb N$, and $x_n \to l$ and $z_n \to l$ for some $l \in \mathbb R$, then $y_n \to l$ also.

So we have to use the definition of convergence to a limit for a sequence: $$\forall \varepsilon > 0, \space \exists N_\varepsilon \in \mathbb N, \space \forall n \ge N_\varepsilon, \space |a_n - l| < \varepsilon$$

I've been trying to say something like: $|y_n - l| < |x_n - l| + |z_n - l| \le \frac\varepsilon 2 + \frac\varepsilon 2 = \varepsilon$ for every $\varepsilon > 0$, but I'm not sure how to get there or if there may be a better way to prove the theorem. Any help would be greatly appreciated.

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?