I have to show that:
If $K=\overline{K}$ and $ X \subseteq K^n $ and $ X $ is open and closed, then $ X = K^n $ or $ X = \emptyset $
Consider $ X = Z (f_1, ..., f_r) = K^n - Z (g_1, ..., g_t) $, then $ Z (f_1, ..., f_r) = K^n \cap Z (g_1, ..., g_t)^c $ and $ Z (f_1, ..., f_r) \cup Z (g_1, ..., g_t) = (K^n \cap Z (g_1, ... , g_t)^c) \cup Z (g_1, ..., g_t) = K^n $ then $ Z (<f_1, ..., f_r> <g_1, ..., g_t>) = K^n $, i.e., $ Z (f_1 g_1, ..., f_1 g_t, ..., f_r g_t) = K^n $, then $ 1 \in <f_1 g_1, ..., f_1 g_t, ..., f_r g_t> $
is what I have, but I do not know how to conclude that $ X = \emptyset $ or $ X = K^n $
Suppose that every non zero polynomial in $K$ does not induces the trivial function. $K^n=Z(f_1g_1,...,f_rg_t)$ implies that $(f_1g_1,...,f_rg_t)=0$. This implies that $(f_1,...,f_r)=0$ or $(g_1,...,g_t)=0$. We deduce that $X=K^n$ or $X$ is empty.