Support of a sheaf need not be closed

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To prove that the support of a sheaf is not necessarily closed I consider this sheaf:

$\mathcal{F}:=\oplus_{p_i \in [0,1)}\mathrm{Sky}_{p_i}\mathbb{Z}$.

Then we have that $\mathrm{Supp}(\mathcal{F})=[0,1) \subset \mathbb{R}$ which is not closed when we consider the Euclidean topology on $\mathbb{R}$.

Is my example correct?

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The crucial question is: What are the stalks of $\mathcal{F}$? For any open $U \subset \mathbb{R}$, one has $$\mathcal{F}(U) = \bigoplus_{x \in U \cap [0,1)} x \cdot \mathbb{Z},$$ and for any inclusion $V \subset U$, the restriction maps $\mathcal{F}(U) \to \mathcal{F}(V)$ factors out all summands that do not belong to $V$. So let $\overline f \in \mathcal{F}_x$ be a germ, represented by a section $(U, f)$. Then $f$ only has finitely many entries, say $f = f_x \cdot x + f_1 \cdot y_1 + \dotsb + f_n \cdot y_n$. $V = U \setminus \{y_1, \dotsc, y_n\}$ is an open set, over which $f = f_x \cdot x$. So the germ $\overline{f}$ is fully determined by its value $f_x$, but $f_x = 0$ if $x \notin [0,1)$.

So I think yes, you are right.