Continuous functions and infinity

Yes, in fact, $$\inf_{\delta>0}\ \liminf_{n\to\infty}f(n\delta) =\liminf_{x\to+\infty}f(x).$$ Assuming w.l.o.g. $\liminf_{x\to+\infty}f(x)<\alpha<+\infty$, the open set $A=\{f<\alpha\}$ is unbounded. Therefore, for any non-empty open interval $(a,b)\subset\mathbb{R}_+$ and any $n\in\mathbb{N}$, the set $\cup_{k> n}(ka,kb)$, that contains a right-unbounded interval, meets $A$. Equivalently, for any $n\in\mathbb{N}$, the open set $ B_n:=\cup_{k> n}{1\over k}A$ meets $(a,b)$, so that $B_n$ is dense in $\mathbb{R}_+$. By the Baire category theorem $\cap_{n\ge0}B_n$ is not empty, actually dense, meaning that there exist $\delta>0$ such that $n\delta\in A$ for infinitely many $n$, an this means $\liminf_{n\to\infty}f(n\delta)\le\alpha.$ Being $\alpha$ arbitrary, the claim follows.

Yes, this is true and well known. One of the references I know is the problem book of B. Makarov, M. Goluzina, A. Lodkin and A. Podkorytov (Selected problems in real analysis, Translations of Mathematical Monographs 107, AMS 1992), problem II.1.25 (it is slightly different, but essentially the same).

An idea is that if, on the contrary, there are intervals $\Delta_k=[x_k,y_k]$ onto which $f$ is bounded, $x_k\to \infty$, then we may construct a nested family of closed intervals $T_i$, $T_1\supset T_2\supset T_3\dots$, and positive integers $n_1<n_2<\dots$, such that each interval of the form $n_iT_i$ is contained in some $\Delta_k$. Then for a common point $t=\cap T_i$ it appears that $f(nt)$ does not tend to infinity.