= Solution
A necessary qualification is $\sigma^2>0$. As printed, the assumptions also permit $X_i=0$ almost surely; then $\mathbb P(A_n)=0$ for every $K>0$, so the asserted positive lower bound is false. Assume the intended nondegenerate case.
The <central limit theorem> states that independent identically distributed variables with mean zero and finite positive variance satisfy $S_n/(\sigma\sqrt n)\Rightarrow Z$, where $Z\sim N(0,1)$. Since the <normal distribution> has no atom at $K/\sigma$,
$$
\mathbb P(A_n)\longrightarrow\mathbb P(Z\geq K/\sigma)=1-\Phi(K/\sigma)>0.
$$
For example,
$$
\boxed{c=\tfrac12[1-\Phi(K/\sigma)]\text{ works for all sufficiently large }n.}
$$
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