Prove $f$ has a maximum on a continuous $f: [0, \infty) \to [0, \infty)$ if $\lim\limits_{x \to \infty} f(x) = 0$.
So $\lim\limits_{x \to \infty} f(x) = 0$ means that for all $\epsilon > 0$ there exists an $R$ such that $0 \leq f(x) < \epsilon$ for all $x \geq R$.
Case 1: If $f(x) = 0$ for all $x \in [0, \infty)$, then $R = 0$ so $0 \leq f(x) < \epsilon$ for all $x \in [0, \infty)$ and for all $\epsilon > 0$, and so the maximum is just $0$.
Case 2: There is at least one $c \in [0, \infty)$ such that $f(c) > 0$, so for all $\epsilon > 0$ there is an $R \neq 0$ such that $0 \leq f(x) < \epsilon$ for all $x \geq R$. At this point intuitively I know there should be a maximum since $f$ is continuous and $\lim\limits_{x \to \infty} f(x) = 0$ at some point after $x = 0$, but I don't really know how to rigorously prove this.
This is not that difficult. Let's assume that that $f(x)$ is not identically $0$ otherwise $0$ is the maximum value of $f(x)$. Then as you mention in your post that there are numbers $a, c$ such that $a \geq 0, c > 0$ such that $f(a) = c$. Now we also know that $\lim\limits_{x \to \infty}f(x) = 0$. Hence there is a number $N > 0$ such that $f(x) < c$ for all $x \geq N$. Moreover we can always choose $N > a$.
Thus we have $f(x) < c$ for $x \geq N$ and $f(x) = c$ for at least one value $a \in [0, N]$. Let $M$ be the maximum value of $f(x)$ in $[0, N]$. Clearly this is guaranteed because $f(x)$ is continuous in closed interval $[0, N]$. It is obvious that $M \geq f(a) = c$. And since $f(x) < c \leq M$ for all $x \geq N$. It follows that $M$ is the maximum value of $f(x)$ on $[0, \infty)$.