Yes, it's possible to have an infinite chain.
I'm sure you're already familiar with some examples:
$$
O(x) \subseteq O(x^2) \subseteq \ldots \subseteq O(x^{42}) \subseteq \ldots
$$
You have an infinite chain here: polynomials of growing degree. Can you go further? Sure! An exponential grows faster (asymptotically speaking) than any polynomial.
$$
O(x) \subseteq O(x^2) \subseteq \ldots \subseteq O(x^{42}) \subseteq \ldots O(e^x)
$$
And of course you can keep going: $O(\mathrm{e}^x) \subseteq O(x\,\mathrm{e}^x) \subseteq O(\mathrm{e}^{2x}) \subseteq O(\mathrm{e}^{\mathrm{e}^x}) \subseteq \ldots$
You can build an infinite chain in the other direction too. If $f = O(g)$ then $\dfrac{1}{g} = O\left(\dfrac{1}{f}\right)$ (sticking to positive functions, since around here we discuss asymptotics of complexity functions). So we have for example:
$$
O(x) \subseteq O(x^2) \subseteq \ldots
\subseteq O\left(\dfrac{e^x}{x^2}\right) \subseteq O\left(\dfrac{e^x}{x}\right) \subseteq O(e^x)
$$
In fact, given any chain of functions $f_1, \ldots, f_n$, you can build a function $f_\infty$ that grows faster than all of them. (I assume the $f_i$'s are functions from $\mathbb{N}$ to $\mathbb{R}_+$.) First, start with the idea $f_\infty(x) = \max \{f_n(x) \mid n \in\mathbb{N}\}$. That may not work because the set $\{f_n(x) \mid n \in\mathbb{N}\}$ can be unbounded. But since we're only intersted in asymptotic growth, it's enough to start small and grow progressively. Take the maximum over a finite number of functions.
$$
f_\infty(x) = \max \{f_n(x) \mid 1 \le n \le N \} \qquad \text{if \(N \le x \lt N+1\)}
$$
Then for any $N$, $f_N \in O(f_\infty)$, since $\forall x \ge N, f_\infty(x) \ge f_N(x)$. If you want a function that grows strictly faster ($f_\infty = o(f_\infty')$), take $f_\infty'(x) = x \cdot (1 + f_\infty(x))$.