Here is an explanation by example. Let $\Theta'(f) = \{ g : g(n)/f(n) \text{ tends to a positive limit} \}$. I will show that $\Theta'(f) \subseteq \Theta(f)$. The first step is to write the definitions. There are several variants of the definitions, and I picked one arbitrarily; you should use the definitions that were stated in class.
Let $f\colon \mathbb{N} \to \mathbb{N}$.
- $\Theta'(f)$ consists of all functions $g\colon \mathbb{N} \to \mathbb{N}$ such that for some positive $x \in \mathbb{R}_+$, $\lim_{n\to\infty} g(n)/f(n) = x$.
- $\Theta(f)$ consists of all functions $g\colon \mathbb{N} \to \mathbb{N}$ such that for some positive $a,b \in \mathbb{R}_+$ and $n_0 \in \mathbb{N}$, for all $n \geq n_0$ it holds that $a f(n) \leq g(n) \leq b f(n)$.
The next step is to take a function $g \in \Theta'(f)$.
Let $g \in \Theta'(f)$. Then there exists a positive $x \in \mathbb{R}_+$ such that $$ \lim_{n\to\infty} \frac{g(n)}{f(n)} = x.$$
Finally, we have to show that $g \in \Theta(f)$. For that we have to come up with positive $a,b \in \mathbb{R}_+$ and $n_0 \in \mathbb{N}$ such that $af(n) \leq g(n) \leq bf(n)$ for all $n \geq n_0$.
By the definition of limit, there exists $n_0$ such that for $n \geq n_0$,
$$ \left|\frac{g(n)}{f(n)}-x\right| \leq \frac{x}{2}. $$
Therefore for $n \geq n_0$,
$$ \frac{x}{2} \leq \frac{g(n)}{f(n)} \leq \frac{3x}{2}, $$
implying
$$ \frac{x}{2} f(n) \leq g(n) \leq \frac{3x}{2} f(n). $$
Taking $a = x/2$ and $b = 3x/2$, we see that $g \in \Theta(f)$.
Only the latter two steps really belong to the proof. The first one states the claim. It is a very important step — without knowing what you're trying to prove, you won't be able to prove it.