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.

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$$|y_n-l|\le |y_n-x_n|+|x_n-l|\le( z_n-x_n)+|x_n-l|$$
Can you take it from here?

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, $y_n \to l$.

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:
&\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
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,
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
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.

How about
where last inequality of the above holds for $n\geq N$ if $N$ is big enough?

An alternative proof is the following:

Proof: Since $x_n \leq y_n \leq z_n$ then $0\leq y_n-x_n\leq z_n-x_n$, thus $|y_n-x_n|\leq z_n-x_n$.

Combining the above with the fact that $\lim(z_n-x_n)=\lim z_n-\lim x_n=l-l=0 \ $, we get: $$\lim(y_n-x_n)=0$$

Now we can write the terms of $(y_n)$ as the sum of the terms of two converging sequences: $y_n=(y_n-x_n)+x_n$, so we have:
\lim y_n=\lim\big((y_n-x_n)+x_n \big)=\lim(y_n-x_n)+\lim x_n=0+l=l

As $n$ grows, you can get the distance from $x_n$ to $l$ to be less than any $\epsilon > 0$, and the distance from $z_n$ to $l$ to be less than $\epsilon$ as well. Just take the max $N$ of the two indices for $x_n$ and $z_n$ that guarantee this, and you will get that both $x_n$ and $z_n$ are within $\epsilon$ of $l$ for $n \geq N$. What does that say about the distance between $y_n$ and $l$, for $n \geq N$?