Question: Suppose that $f(x) \leq g(x) \text{ for all }x$. Prove that $\lim\_{x\to a} f(x) \leq \lim\_{x \to a} g(x)$ provided these limits exist.

Attempt: I am not even sure how to start this question. I had made an attempt but it was getting nowhere when I did it. I found a solution to the question, but I still do not understand what is going on:

Suppose $$ l = \lim\_{x\to a}f(x) \geq \lim\_{x\to a}g(x) = m$$. Let $$\epsilon = l - m > 0$$ 

Then there is $$\delta > 0 \text{ such that if } 0 <|x-a|<\delta \Rightarrow |l-f(x)|< \frac{\epsilon}{2} \text{ and } |m-g(x)| < \frac{\epsilon}{2}$$ 

Thus for $$0 <|x-a|<\delta \text{ we have } g(x) < m + \frac{\epsilon}{2} = l - \frac{\epsilon}{2} < f(x)$$ 

Contradicting the hypothesis.

My problem is I do not see clearly what the hypothesis is and of equal importance I don't see how this proof accomplishes the objectives of the question.

Answer: 
The hypothesis is that $f(x)\le g(x)$ for all $x$. We want to show under this hypothesis that

$$\lim\_{x\to a}f(x)\le\lim\_{x\to a}g(x)\;.\tag{1}$$

The idea of the proof is to show that if $(1)$ is false, then so is the hypothesis: we’ll assume that $(1)$ is false and use that assumption to find an $x$ such that $f(x)\not\le g(x)$, i.e., such that $f(x)>g(x)$.

The first step is just for convenience, so that we don’t have to write so much: we let $\ell=\lim\_{x\to a}f(x)$ and $m=\lim\_{x\to a}g(x)$. Now we assume that $(1)$ is false, i.e., that $\ell>m$. (You have $\ell\ge m$ in your question, but that can’t be right.) Then we let $\epsilon=\ell-m$; since we’re assuming that $\ell>m$, we know that $\epsilon>0$.

Now we apply the definition of limit twice. Since $\ell=\lim\_{x\to a}f(x)$, there is a $\delta\_f>0$ such that $|\ell-f(x)|<\frac{\epsilon}2$ whenever $0<|x-a|<\delta\_f$. And since $m=\lim\_{x\to a}g(x)$, there is a $\delta\_g>0$ such that $|m-g(x)|<\frac{\epsilon}2$ whenever $0<|x-a|<\delta\_g$. Now let $\delta=\min\{\delta\_f,\delta\_g\}$; clearly

$$|\ell-f(x)|<\frac{\epsilon}2\quad\textbf{and}\quad|m-g(x)|<\frac{\epsilon}2$$

whenever $0<|x-a|<\delta$. The first of these implies that

$$f(x)>\ell-\frac{\epsilon}2$$

whenever $0<|x-a|<\delta$, and the second implies that

$$g(x)<m+\frac{\epsilon}2$$

whenever $0<|x-a|<\delta$. Put the two together: if $0<|x-a|<\delta$, then

$$g(x)<m+\frac{\epsilon}2=\ell-\frac{\epsilon}2<f(x)\;.$$

We’ve found not just *one* $x$ such that $g(x)>g(x)$: we’ve found a whole interval of them. This certainly contradicts the hyposthesis that $f(x)\le g(x)$ for all $x$, and that contradiction shows that our assumption that $\ell>m$ is impossible. Thus, it must be true that $\ell\le m$, which is what we wanted to prove.
Your proof by contradiction should begin with a strict inequality: "Suppose $l = \lim\_{x\to a}f(x) > \lim\_{x\to a}g(x) = m$." Continuing:

Pick an $\epsilon > 0$ so small that the neighborhoods $Y\_{l} = ]l - \epsilon, l+\epsilon[, Y\_{m} = ]m - \epsilon, m+\epsilon[$ have no point in common. (**Note**: $l - \epsilon > m + \epsilon$.) There exist: (1) a neighborhood $X\_{f}$ of $a$ that gets mapped by $f$ into $Y\_{l}$, and (2) a neighborhood $X\_{g}$ of $a$ that gets mapped by $g$ into $Y\_{m}$. Therefore, every point $x$ in $X\_{f} \cap X\_{g}$ gets mapped into $Y\_{l}$ by $f$ and into $Y\_{m}$ by $g$. But then, by the **Note** above, $f(x) > g(x)$. This is a contradiction.