Is this a valid proof of the squeeze theorem? I'm self-studying Spivak's calculus, and I have no way of checking my solutions. One of the problems asks for a proof of the squeeze theorem. Here is what I have figured: 
Proof. Suppose there exist two functions $f(x)$ and $g(x)$ such that $(\forall x)(g(x) \geq f(x))$ and both $\lim _{x\to a}f(x) = l_1$ and $\lim _{x\to a}g(x) = l_2$ exist. Now let $h(x) = g(x) - f(x)$ so that $h(x) \geq 0$. Assume that $l_2$ $<$ $l_1$. 
By previous results, $(\forall \epsilon > 0)(\exists \delta > 0): (\forall x)($if $\lvert x - a \rvert < \delta$, then $\lvert h(x) - (l_2 - l_1) \rvert < \epsilon)$. 
But if $\epsilon \leq l_1 - l_2$, no $\delta$ exists such that $\lvert h(x) - (l_2 - l_1) \rvert < \epsilon$, thus, a contradiction. 
$\therefore l_2 \geq l_1$ 
Given this result, define three functions $f_1(x)$, $f_2(x)$, and $f_3(x)$ such that $(\forall x)(f_1(x) \leq f_2(x) \leq f_3(x))$ and $\lim _{x \to a}f_1(x) = \lim _{x \to a}f_3(x) = L$. 
Because $f_2(x) \geq f_1(x)$, it must be true that $\lim_{x \to a}f_2(x) \geq \lim_{x \to a}f_1(x)$ and similarly, because $f_3(x) \geq f_2(x)$, it must also be true that $\lim_{x \to a}f_3(x) \geq \lim_{x \to a}f_2(x)$. 
It follows that $\lim_{x \to a}f_1(x) \leq \lim_{x \to a} f_2(x) \leq \lim_{x \to a} f_3(x)$, which in turn makes true that $L \leq lim_{x \to a} f_2(x) \leq L$. 
$\therefore \lim_{x \to a}f_2(x) = L$ $\blacksquare$
I'm mostly unsure about the validity of the first part, specifically the "no $\delta$ exists" bit. Also, I would appreciate suggestions regarding formal proof-writing as I'm new to this. Thanks a lot in advance! 

Why does no such $\delta$ exist? This is not immediately obvious. you need to explain it.
A second problem with your proof is that the squeeze theorem only assumes that $f_1$ and $f_3$ converge. That the squeezed function $f_2$ also has a limit is part of what must be proved. But your proof simply assumes this is so.
For a better approach, consider this: if $\epsilon > 0$, then there is a $\delta_1 > 0$ such that for $|x - a| < \delta_1$, $f_1(x) > L - \epsilon$. And there is a $\delta_2 > 0$ such that for $|x - a| < \delta_2$, $f_3(x) < L + \epsilon$.
