I think I got an conundrum that I put myself with the wrong image of etale sheaf. I have finished the problem by Hartshorne's version but when I transfer this problem to the statement of etale sheaf, I got a completely different statement. This two statement should be completely equivalent due to that etale sheafification is an isomorphism on sheaf category as a subcategory of presheaf category to etale sheaf category.
This is related to Hartshorne 2.1.14 and 2.1.13.
Let $F$ be a sheaf over $X$. Let $U\subset X$ open and $s\in F(U)$. Equivalently, I could think $s$ as a continuous section from $U\to F|_U$ where $F|_U$ is the etale sheaf restricted to $U$. So locally it should be pictorially like a vector bundle but with fiber discrete topology. I want to show $Supp(s)$ is closed. However my argument shows that $Supp(s)$ is also open.
If $s|_p=0$, then clearly, I have a continuous section locally agree with $0$ section locally on an open set. In particular, from continuity of continuous section $s$ on etale, $s(\varinjlim_ix_i)=\varinjlim_i(s(x_i))$. If I have a sequence of points $x_i$ converging to $p$ and their $s(x_i)=0$, then I must have $s(p)=0$ by continuity.
Thus any section not vanishing on $p$ must not-vanishing locally in some open set around $p$. So for any $p\in Supp(s)$, I found a small open set on which it does not vanish. So I conclude $Supp(s)$ is open? If this is the case, then I have $X$ disconnected. There is no reason to expect this.
The closedness can be shown as the following. $Supp(s)^c$ is open as I can find a small section vanishing locally containing each point of $Supp(s)^c$.
Q1. Which part have I done wrong in showing openness of $Supp(s)$? I hope someone could point out whether my image of etale sheaf is correct.
Q2. How should I properly visualize etale sheaf?
First of all, do not use (in the present context) the notation $s\vert_p$.
If you want to consider a sheaf $\mathcal F$ as an étalé space, then a section of that sheaf over the open subset $U\subset X$ is a continuous map $U\to \mathcal F$ which composed with the projection $p:\mathcal F\to X$ yields the inclusion map $p\circ s:U\hookrightarrow X$.
It thus makes sense to use the notation $s(p)\in \mathcal F$, in conformity with the general set-theoretic notion of "value of a function at a point".
It is then perfectly clear that the support of $s$ is closed since its complement $Supp(s)^c$ is open: if $s(p)=0$, then $s(q)$ will be zero too for all $q$ in a neighbourhood $U(p)$ of $p$.
However the support of $s$ has no reason to be open:
For example if $\mathcal C$ is the sheaf of continuous real-valued functions on $\mathbb R$, the function $f$ defined by $f(x)=0$ for $x\leq 0$ and $f(x)=x$ for $x\gt 0$ has support $[0, \infty )$ when seen as a section of the sheaf $\mathcal C$, and the good news is that this coincides with the support of $f$ as defined in elementary calculus.