Can I "close" an open interval $[0,\infty)$ as $x$ approaches infinity with some real number, if given that $\displaystyle \lim_{x \to \infty }f(x)=f(0)$ ?
The final goal of the exercise is to prove that $f$ isn't one-to-one. So I thought I could use Weierstrass theorem to prove that a bounded interval (our new interval $[0,M]$ where $M$ as our "infinity" constant) has min/max so it will definitely will have $x_1,x_2$ that will $f(x_1) = f(x_2)$
First of all, you can do that as long as you also prove that the theorem holds in that context. I don't think that that's worth the trouble.
Even if you do that, I don't see how is it that you deduce from that there there are $x_1,x_2\in[0,\infty)$ such that $x_1\ne x_2$ and that $f(x_1)=f(x_2)$.
Anyway, if $f$ is constant, it is obvious that it is not injective. If there is some $x_0\in[0,\infty)$ such that $f(x_0)>f(0)$, then, since $\lim_{x\to\infty}f(x)=f(0)$, if $x\gg0$ you have $f(x)<f(x_0)$. So, take some $x_1>x_0$ such that $f(x_1)<f(x_0)$. If $f(x_1)\geqslant f(0)$, there is some $x_2<x_0$ such that $f(x_2)=f(x_1)$. ANd if $f(x_1)<f(0)$, ther is some $x_2\in(x_0,x_1)$ such that $f(x_2)=f(0)$.
The case in which there is some $x_0\in[0,\infty)$ such that $f(x_0)<f(0)$ is similar.