I guess the argument you mention is the following:
If $\dim_{\mathbb C} X = 1$ the statement is tautological. Assume that $\dim X = d > 1$ and $X \hookrightarrow \mathbb{P}^n$ is projective. Choose any two points $x, y \in X$. Then there exists a projective subspace $H \subset \mathbb{P}^n$ of dimension $n-d+1$ which passes through both $x$ and $y$.
Consider $S := H \cap X$. If $\dim S = 1$, we are done. Otherwise do the same trick with $S \hookrightarrow H$ itself and use the induction by the dimension.
Clearly the same argument works with quasi-projective varities. The only specification is that the curve you've constructed does not have to be complete. Moreover there is a case when it never does: take a projective $\overline{X}$ and $X := \overline{X} \setminus D$, where $D$ is a hyperplane section divisor. Then any curve on $\overline X$ will intersect the hyperplane which cuts out the $D$ and therefore will intersect $D$ itself.
I thing in a general case of algebraic variety you can try to find an affine (and hence quasi-projective) chart which would contain both of your points $x$ and $y$, find a quasi-projective curve on this chart and then take its closure in the ambient variety (I suspect that the separateness and the finite-type conditions allow you to do this).
After all, I have to mentioned that I have never met a complex algebraic variety which is not quasi-projective. This seemes to be something rather exotic. The accurate generalization of complex projective varities is a notion of Kähler manifold. For arbitrary Kähler manifold the property you're asking about fails.