Zariski dense implies classically dense? I was surprised that I wasn't able to find this question already posted; if it has been posted and I just didn't find the right search terms, let me know.
Let $X$ be any complex variety. A priori, any set which is dense in the classical topology on $X$ is automatically dense in the Zariski topology on $X$, just because the Zariski topology has fewer open/closed sets.
For the converse, this question, though distinct, does shed some light: in $\mathbb A_{\mathbb C}^1$, any infinite subset is Zariski dense, but certainly not necessarily classically dense.
But in my experience, it seems that any Zariski open subset of $X$ that is Zariski dense is also classically dense. Is this true? How do you prove it?
 A: Let $T$ be a locally constructible subset of a finite type $\mathbb C$-scheme $X$. (You can take $T$ to be a (Zariski) open of a complex algebraic variety, for instance.)
Then $T$ is dense in $X$ if and only if $T(\mathbb C)$ is dense in $X(\mathbb C)$. A reference for this is Expose XII, Cor. 2.3 p. 243 of SGA 1. 
http://arxiv.org/pdf/math/0206203v2.pdf
A: Let's start with the  affine case:
Claim. If $A\subsetneq\mathbb A^n_{\mathbb C}$ is algebraic, then its complement is dense.
Proof.
Let $A=\{\,x\in\mathbb A^n\mid \forall f\in S\colon f(x)=0\,\}$ where the $S\subseteq \mathbb C[X_1,\ldots, X_n]$.
For $a\in A$ we have to exhibit points close to $a$ that are $\notin A$.
Pick $b\in\mathbb A^n\setminus  A$ and
$f\in S$ with $f(b)\ne 0$.
Then the polynomial $g(T)=f(a+(b-a)T)\in\mathbb C[T]$ is not the zero polynomial.  Hence its root at $0$ is isolated and  so $g(h)\ne 0$ for all sufficiently small nonzero $h$. 
Then $a+h(b-a)\notin A$ for such $h$, showing the claim. $_\square$
The extension to the projective case and then to the locally quasiprojective case  should be clear
A: I faced recently the same problem (see this question) and found a complete proof in Mumford's book Algebraic Geometry I, Complex Projective Varieties on page 38, that's Theorem 2.33.
