Question about a Symmetric random walk, Problem 4.1.1 in Durrett I am working on the following problem:

Let $X_1, X_2, \dots \in \mathbb{R}$ be i.i.d. with a distribution that is symmetric about $0$ and nondegenerate, i.e. $P(X_i=0)<1$. Show that $-\infty = \liminf S_n < \limsup S_n=\infty$. Here $S_n=X_1 + \dots +X_n$.

I know that my two events are exchangeable and so by a corollary of the Hewitt-Savage $0$-$1$ law one of the following events occurs with probability one:
$
i) \ S_n=0 \text{ for all } n,
$
$
ii) \ S_n\rightarrow \infty,
$
$
iii) \ S_n\rightarrow -\infty,
$
$
iv) -\infty = \liminf S_n < \limsup S_n=\infty.
$
Since $P(X_i=0)<1$, $(i)$ cannot occur. I am pretty sure that if $(ii)$ holds then $(iii)$ must hold by the symmetry of the distribution and vice versa, which would be a contradiction. However, I cannot think of a mathematically precise way to say this.
Any help is appreciated!

If $S_n\to\infty$ a.s. then for $\epsilon>0$ there is $N$ s.t. for $n>N$, $P[S_n>1]>1-\epsilon$, say. But you know for any finite $n$, $P[S_n<0]=P[S_n>0]$ (E.g. induction. Given symmetric independent RVs $U$ and $V$, $P[U+V<0]=\int_x F_U(-x)dF_V(x)=\int_x(1-F_U(x))dF_V(-x)=P[U+V>0]$, ignoring continuity issues.)
