Prove that $f : [-1, 1] \rightarrow \mathbb{R}$, $x \mapsto x^2 + 3x + 2$ is strictly increasing. 
Prove that $f : [-1, 1] \rightarrow \mathbb{R}$, $x \mapsto x^2 + 3x + 2$ is strictly increasing.

I do not have use derivatives, so I decided to apply the definition of being a strictly increasing function, which should be: 

If we pick 2 numbers $a$ and $b$ from the domain of a function $f$, where $a < b$, then $f(a) < f(b)$.

Now (if that is a correct definition), I have tried to apply it to my case:

Let $a$, $b \in [-1, 1]$ and $a < b$. We want to show that $f(a) < f(b)$ or that $f(b) - f(a) > 0$.
We know that $x^2 + 3x + 2 = (x + 2)(x + 1)$, thus we have that $f(a) = (a + 2)(a + 1)$ and $f(b) = (b + 2)(b + 1)$ , therefore we need to show that:
$$(b + 2)(b + 1) - (a + 2)(a + 1) > 0$$
We can see that $(b + 2)$ and $(a + 2)$ will always be positive, and that $(b + 2) > (a + 2)$ , since $b > a$ (by assumption).
Since $b > a$, we know that $b > -1$ (otherwise $a \geq b$), thus $(b + 1) > 0$ (so we know that $(b + 2)(b + 1) > 0$ .  We also have at most $(a + 1) = 0$, and thus  we have that $(a + 2)(a + 1) \ge 0$.
So, here is the proof that  $f(b) - f(a) > 0$ .  

Am I correct? If yes, what can I improve it? If not, where are the erros and possible solutions?

The proof given looks fine to me, modulo my comment (see above).
Here's another way to see it, sans calculus:
Choose $x_1, x_2 \in [-1, 1]$ with $x_1 > x_2$.  Then
$f(x_1) - f(x_2) = x_1^2 + 3x_1 + 2 - x_2^2 - 3x_2 - 2 = x_1^2 - x_2^2 + 3(x_1 - x_2)$
$= (x_1 + x_2)(x_1 - x_2) + 3(x_1 - x_2) = (x_1 + x_2 + 3)(x_1 - x_2); \tag{1}$
note that $x_1 - x_2 > 0$ since $x_1 > x_2$ and that $x_1 + x_2 + 3 > 0$ since $x_1 + x_2 > -2$ since $x_1 > x_2 \ge -1$.  Thus
$f(x_1) -f(x_2) > 0, \tag{2}$
or
$f(x_1) > f(x_2), \tag{3}$
and we're done!
Note that all we really needed was $x_1 > x_2 \ge -1$.
Of course, with calculus we see that $f'(x) = 2x + 3 >0$ for $x \ge -1$, etc. etc. etc.  And since the minimum occurs at $x = -3/2$, Mark Bennet's comment is corroborated.
