*You need $U$ "contractible"*. In general $\Delta(S)$ is defined exactly as in the $1$-topos case: "take the constant presheaf valued at $S$ and sheafify it".

This formula you are giving involving connected open is not a definition but a 'theorem', and for example you could have a space or topos where no $U$ is connected, think about sheaves over the Cantor space for example.

This theorem is also true for $\infty$-topos, but as far as I'm concerned this is the definition of a contractible $\infty$-topos (to put in another way, if you want to see a proof it, it highly depends on what you mean by "U is contractible" ).


To give an example that should show you that connected is not enough, consider sheaves on a nice space $X$ (typically a CW-complex), and take $S$ to be some Eilenberg-Mac lane space $S = K(\pi,n)$. Then in general, $\Delta(S)(U)$ for $U \subset X$ an open subspace is the space of maps from $U$ to $S$.

So as soon as as $H^n(U,\pi)$ is non zero, $\Delta(S)(U)$ has several connected component (corresponding exactly to elements of $H^n(U,\pi)$) while $S$ itself only has one connected component.

Note that in this special case (when $X$ is a nice topological space) it is easy to see that the formula $\Delta(S)(U) = S$ holds for all $S$ exactly when $U$ is contractible as a topological space in the usual sense. But, as mentioned before, for a more general topos you might want to clarify what "contractible" should mean in the first place...