Proving that $f:\mathbb{R}_0^+ \to \mathbb{R},~f(x) = x^2 + 4x + 4$ is injective using direct proof from definition 
Given function $f:\mathbb{R}_0^+ \to \mathbb{R},~f(x) = x^2 + 4x + 4$ prove that it is injective.

Using definition of injectivity $(\forall x_1, x_2 \in \mathbb{R}_0^+)(x_1 \neq x_2 \implies f(x_1) \neq f(x_2))$ I'm doing the following:
$$x_1^2 + 4x_1 + 4 = x_2^2 + 4x_2 + 4$$
$$x_1^2 - x_2^2 = -4(x_1 - x_2)$$
$$x_1 + x_2 = -4$$
$$x_1 = -4 - x_2.$$
Since domain is $\mathbb{R}_0^+$ it is apparent that $x_1 \neq -4 - x_2$ and hence function is not injective.

Is my final argument correct? In cases like that, shall I use definition instead of counterpositive?
 A: Your argument is fine, but you can end a step earlier: $x_1>0$ and $x_2>0$ implies $x_1+x_2>0$, so $x_1+x_2=-4$ is a contradiction.
However you have better making evident where you're using the hypothesis $x_1\ne x_2$.

Proofs of injectivity are often times simpler with the contrapositive: “if $f(x_1)=f(x_2$ then $x_1=x_2$”.
Suppose $f(x_1)=f(x_2)$; then, as in your steps,
$$
(x_1^2-x_2^2)+4(x_1-x_2)=0
$$
so
$$
(x_1-x_2)(x_1+x_2+4)=0
$$
Since $x_1+x_2+4>x_1>0$, we conclude $x_1-x_2=0$.
A: Your final argument is not correct.
Instead you have (from your calculation):
Since $x_2 \geq 0$ that would imply $x_1 <0$ which is impossible (because out of the domain): therefore f is injective on the domain.
It would also be better to give your assumptions: suppose there exists $x_1 \neq x_2$ such that $f( x_1)=f(x_2)$...
A: $\exists (x_1,x_2) \in [0,+\infty)^2 : x_1^2+4x_1+4=x_2^2+4x_2+4 \; \implies$
$(x_1-x_2)(x_1+x_2+4)=0 \implies$
$x_1=x_2 $ since $x_1+x_2+4\geq4$.
thus $f$ is injective.
A: You have $$f(x)=(x+2)^2$$ so if the function has domain $\Bbb R$ then the points
$x=t-2$ and $x=-t-2$ have same image for all $t$ so the function $f$ could not be injective. However with domain $\mathbb{R}_0^+ $ it is in fact injective and strictly increasing on its domain because is of the quadratic form $x^2$.
