Consider $F\in\operatorname{Shv}(X, Pr^L)$, we show that $F$ is hypercomplete, i.e., it satisfies hyperdescent.

For any $Z \in Pr^L$, the assignment

\[h_X(Z): U \mapsto \operatorname{map}(Z, F(U))\]

is a sheaf of spaces because taking limit commutes with taking mapping spaces (in the second variable).

Note that sheaves of spaces on a manifold are hypercomplete, because it has finite homotopy dimension, which in turn comes from the fact that it has finite covering dimension, see HTT 7.2.

Therefore, $h_X(Z)$ satifisfies hyperdescent. By Yoneda lemma, so does $F$.

Note that this argument carries actually to sheaves with coefficients in any $\infty$-category. Thus the topos of sheaves of spaces is universal in this sense.