Let $X$ be a topological space. Let $I$ be a poset and let $\mathcal F_i$ for $i\in I$ be sheaves on $X$, and $\{\pi_{ij}\colon \mathcal F_i \rightarrow \mathcal F_j\}_{i,j\in I}$ be an inductive system of maps for $i \le j$ and $i,j \in I$ that are compatible with the restriction maps of the $\mathcal F_i$. One may form a sheaf $\varinjlim \mathcal F_i$ by sheafifying the presheaf $U\mapsto \varinjlim \mathcal{F}_i(U)$ for $U$ open in $X$.
Claim: If $X$ is Noetherian then the presheaf $U\mapsto \varinjlim \mathcal{F}_i(U)$ is already a sheaf.
The above is an exercise in Hartshorne. It seems to me that this is true more generally, indeed all one needs is the space to be quasi-compact. But being Noetherian is stronger, indeed one can prove that a space is Noetherian iff it is locally Noetherian and quasi-compact. So I am left wondering why Mr. Hartshorne wrote Noetherian instead of quasi-compact here? Perhaps I am missing something?
Be aware that the sheaf axioms have to hold for any open covering of any open subset of $X$. I have not checked but, for me, that strongly suggests that any open subset of $X$ must be quasi-compact - which is equivalent to the space being Noetherian.
(Comment posted as an answer on @sdf's suggestion.)