Let $q: W \to X$ be a proper birational morphism of quasi-projective varieties. Let $E$ be an effective, $q$-exceptional divisor. Suppose $C \subseteq E$ is a curve, then is it true that the intersection $C \cdot E < 0$?
I had a very vague idea to "prove" such "result" (and I think this is somehow the standard way) : cut $X$ by hypersufaces and prove the result for surfaces using Hodge index theorem. However, I am not very confident with how to do the induction step...neither do I know the validity of the result.
(I'm sorry, this should be a comment, but I cannot comment yet). I am not sure your claim is true.
Take a quartic surface $X \subset \mathbb P^3$ with a singularity $p$ of type $A_3$ and let $W$ be a minimal resolution. $W$ is a K3 surface and the fibre over $p$ is a divisor $E$ with three rational irreducible components $C_1$, $C_2$ and $C_3$, such that $C_i \cdot C_{i+1}=1$ for $i=1,2$ and $C_i^2 = -2$.
If you take $C = C_2$ then $C \cdot E = 1 -2 +1 = 0$.