When is it true that if a section vanishes on a dense open set then it is the zero global section? In other words, when do global sections act like continuous functions? Is it true on an integral scheme?
I am asking this question because I am wondering whether it is possible for a non-zero global section on an integral scheme $(X,O_X)$ to vanish on a non-empty set of stalks. I mean, if $s \in O_X(X)$ with $s_x=0 \in O_{X,x}$ for some $x$ then $s$ vanishes on some open set which is dense, and if my claim is true then it is zero. This would imply that for any $s \in O_X(X),$ ($s_x=0$ for some $x$) $\Rightarrow(s = 0)$. Is this true?
If $X = \operatorname{Spec}A$ for $A$ some integral ring, then a non-zero global section is never going to vanish at some stalk since $A$ integral means $A$ contains no non-zero zero-divisors, so $A$ injects in $A_p$. I'm assuming $A= O_X(X) \rightarrow O_{X,x} = A_x$ is the canonical injection. I don't know actually...
If $X$ is integral then I guess $X$ can be covered with open affine subsets $\operatorname{Spec}A_i$ with $A_i$ integral. If I take a non-zero global section then its restriction $s_i$ to some $\operatorname{Spec}A_i$ is also going to be non-zero everywhere on $\operatorname{Spec}A_i$ which is dense. So $s$ would be non-zero on a dense open subset. So I guess it cannot also vanish at some stalk, else it would also vanish on a dense open subset which is a contradiction since two dense open subsets always intersect non-trivially, is it sound?
 A: 
I am asking this question because I am wondering whether it is possible for a non-zero global section on an integral scheme $(X,O_X)$ to vanish on a non-empty set of stalks. I mean, if $s \in O_X(X)$ with $s_x=0 \in O_{X,x}$ for some $x$ then $s$ vanishes on some open set which is dense, and if my claim is true then it is zero. This would imply that for any $s \in O_X(X),$ ($s_x=0$ for some $x$) $\Rightarrow(s = 0)$. Is this true?

Yes.

If $X = \operatorname{Spec}A$ for $A$ some integral ring, then a non-zero global section is never going to vanish at some stalk since $A$ integral means $A$ contains no non-zero zero-divisors, so $A$ injects in $A_p$. I'm assuming $A= O_X(X) \rightarrow O_{X,x} = A_x$ is the canonical injection. I don't know actually...

Yes, the map called "taking stalks" is the canonical injection $A\to A_p$. This reasoning is perfectly correct.

If $X$ is integral then I guess $X$ can be covered with open affine subsets $\operatorname{Spec}A_i$ with $A_i$ integral. If I take a non-zero global section then its restriction $s_i$ to some $\operatorname{Spec}A_i$ is also going to be non-zero everywhere on $\operatorname{Spec}A_i$ which is dense. So $s$ would be non-zero on a dense open subset. So I guess it cannot also vanish at some stalk, else it would also vanish on a dense open subset which is a contradiction since two dense open subsets always intersect non-trivially, is it sound?

This proof is essentially correct. There are some slight adjustments that can be made here to make this clearer. Suppose there is a nonzero global section $s\in \mathcal{O}_X(X)$. Then there exists a nonempty affine open $U$ so that $s|_U$ is nonzero: if not, then $s=0$ as the affine opens form an open covering on which $s$ restricts to zero, so $s=0$ by the sheaf property. Now run the proof from paragraph 2: $s_u\neq 0$ for any $u\in U$. On the other hand, if $s=0$ on some other affine open $U'$, then $s_x=0$ and $s_x\neq 0$ for all $x\in U\cap U'\neq \emptyset$, a contradiction. So $s$ restricts to a nonzero section on every affine open, and by the previous proof is nonzero in every stalk.
