= Solution
Continue with $\sigma^2>0$. For each integer $K\geq1$, let $B_m(K)=\bigcup_{n\geq m}\{S_n/\sqrt n\geq K\}$. These events decrease with $m$, and their probabilities are at least $c_K>0$ by (iii). <Continuity from above of a measure> gives
$$
\mathbb P\left(\frac{S_n}{\sqrt n}\geq K\text{ infinitely often}\right)\geq c_K.
$$
It is important not to assume that this exact threshold event is automatically a <tail event>: a vanishing change of partial sums can affect infinitely many exact comparisons at the threshold. Instead use
$$
L=\limsup_{n\to\infty}\frac{S_n}{\sqrt n}.
$$
For any fixed $m$, subtracting $S_m/\sqrt n\to0$ shows that $L=\limsup_{n\to\infty}(X_{m+1}+\cdots+X_n)/\sqrt n$. Thus $L$ is measurable with respect to every tail sigma-algebra. The event $\{L\geq K\}$ has positive probability by the preceding bound, so <Kolmogorov zero-one law> makes its probability one. Intersecting over all positive integers gives
$$
\boxed{\limsup_{n\to\infty}\frac{S_n}{\sqrt n}=+\infty\quad\text{almost surely}.}
$$
For any specified real $K>0$, this conclusion implies infinitely many $n$ with $S_n/\sqrt n\geq K$, completing both requests. In the excluded zero-variance case the limsup is zero.
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