The statement of the problem is: The function $f: \mathbb{R} \rightarrow [0, \infty)$ is continuous and moreover $\lim_{x \rightarrow -\infty} f(x)=0$ and $\lim_{x \rightarrow \infty} f(x)=0$. Show that $f$ is bounded above and attains it's upper bound.
Now because of the finite limits at infinity, it's easy and well known to show that $f$ is bounded above. The part I'm struggling with is to show that $f$ attains indeed it's upper bound. I sort of tried to repeat the argument that a continuous function on a bounded interval attains it's bounds:
Attempted proof:
So we know $f$ is bounded above by $C>0$. Now we take an $\epsilon>0$ s.t $C>\epsilon >0$. Since $\lim_{x \rightarrow -\infty} f(x)=0$, $\exists M \in\mathbb{N} $ s.t. $f(x)<\epsilon$ $\forall x<M$ and similarly since $\lim_{x \rightarrow \infty} f(x)=0$, $\exists N \in\mathbb{N} $ s.t. $f(x)<\epsilon$ $\forall x>N$. Now we take the closed, bounded interval $[M, N]$ on which we know $f$ attains it's upper bound, done.
Is my proof correct?
No, the proof is not correct. You seem to think that if $f$ attains an upper bound at some $x_0\in[M,N]$, then $(\forall x\in\mathbb{R}):f(x)\leqslant f(x_0)$.
If $f$ is the null, then it attains its upper bound everywhere. Otherwise, let $x_0\in\mathbb R$ be such that $f(x_0)>0$. Then there's some $N>0$ such that $x>N\implies f(x)<f(x_0)$ and there is some $M<0$ such $x<M\implies f(x)<f(x_0)$. You can assume, without loss of generality, the $N\geqslant x_0$ and that $M\leqslant x_0$, which imples that $x_0\in[M,N]$. Take $x_1\in[M,N]$ such that $f(x_1)=\sup f|_{[M,N]}$. Then, yes, $f(x_1)=\sup f$, because outside $[M,N]$ you have $f(x)<f(x_0)\leqslant f(x_1)$.