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 theorem when one takes $U$ contractible is indeed 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 topological space $X$ (typically a CW-complex), and take $S$ to be some Eilenberg-Mac lane space $S = K(\pi,n)$ (though any space would do).

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... and depending on what definition you go for, it makes the question trivial or not.

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" (i.e. applies the left adjoint to the forget full functor from sheaf to presheaf).

This theorem, when one takes $U$ contractible, is indeed 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 topological space $X$ (typically a CW-complex), and take $S$ to be some Eilenberg-Mac lane space $S = K(\pi,n)$ (though any space would do).

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... and depending on what definition you go for, it makes the question trivial or not.

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...

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 theorem when one takes $U$ contractible is indeed 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 topological space $X$ (typically a CW-complex), and take $S$ to be some Eilenberg-Mac lane space $S = K(\pi,n)$ (though any space would do). 

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... and depending on what definition you go for, it makes the question trivial or not.