For an affine variety $X$ with coordinate ring $A$ it is not hard to see that for $g\in A$ the basic open set (or distinguished open set) $$D(g):=\{ P\in X | g(P)\neq 0\}$$ is dense in $X$ if and only if $g$ is not a zero divisor in $A$.
Now let $U\subset X$ be a dense open subset. It should be true that $U$ contains one of the dense $D(g)$. Is this easy to see? Is it possible to prove this without using any scheme theory?
Let $P_1,\ldots,P_r$ be the minimal primes of $A$. Next let $U=X - Z$ with $I \subseteq A$ an ideal of $A$ and $Z=V(I)$. Now as $U$ is dense in $X$ we have $I \not\subseteq P_i$ for all $i$. Therefore $I \not\subseteq P_1 \cup \cdots \cup P_r$ by the prime avoidance lemma. Now $P_1 \cup \cdots \cup P_r$ is the set of zero-divisors of $A$. (More generally the set of zero divisors of an arbitrary noetherian ring is the union of all its associated primes. But here all associated primes are minimal, as $X$ is a variety, therefore $A$ is a reduced ring).
So there is a $f$ in $I$ that is no zero-divisor of $A$. The $D(f)$ is the basic open set sought for.