Proof.
For every integer $m > n$, the $n$-skeleton $\operatorname{sk}_{n}( \Delta ^ m )$ is contained in the boundary $\operatorname{\partial \Delta }^{m}$, and therefore coincides with the $n$-skeleton of $\operatorname{\partial \Delta }^ m$. We therefore have a commutative diagram of restriction maps
\[ \xymatrix@R =50pt@C=50pt{ \operatorname{Hom}_{\operatorname{Set_{\Delta }}}(\Delta ^ m, X ) \ar [dr]_{ \theta _{ \Delta ^ m} } \ar [rr] & & \operatorname{Hom}_{\operatorname{Set_{\Delta }}}( \operatorname{\partial \Delta }^ m, X) \ar [dl]^{ \theta _{ \operatorname{\partial \Delta }^ m} } \\ & \operatorname{Hom}_{\operatorname{Set_{\Delta }}}( \operatorname{sk}_{n}(\Delta ^ m), X). & } \]
If condition $(1)$ is satisfied, then the vertical maps are injective, so the upper horizontal map is also injective. Moreover, if $m \geq n+2$, then $\operatorname{\partial \Delta }^{m}$ contains the $(n+1)$-skeleton of $\Delta ^ m$. In this case, condition $(2)$ guarantees that the vertical maps have the same image, so that the horizontal map is bijective. It follows that $X$ is weakly $n$-coskeletal.
We now prove the converse. Assume that $X$ is weakly $n$-coskeletal, and let $S$ be a simplicial set. Then we can identify $\operatorname{Hom}_{\operatorname{Set_{\Delta }}}(S, X)$ with the inverse limit of the tower of restriction maps
\[ \cdots \rightarrow \operatorname{Hom}_{\operatorname{Set_{\Delta }}}( \operatorname{sk}_{n+2}(S), X) \rightarrow \operatorname{Hom}_{\operatorname{Set_{\Delta }}}( \operatorname{sk}_{n+1}(S), X) \rightarrow \operatorname{Hom}_{\operatorname{Set_{\Delta }}}( \operatorname{sk}_{n}(S), X). \]
Consequently, to prove $(1)$, it will suffice to show that the restriction map
\[ \operatorname{Hom}_{\operatorname{Set_{\Delta }}}( \operatorname{sk}_{m+1}(S), X) \rightarrow \operatorname{Hom}_{\operatorname{Set_{\Delta }}}( \operatorname{sk}_{m}(S), X) \]
is injective for $m=n$ and bijective for $m > n$. Using Proposition 1.1.4.12, we can reduce to the case $S = \Delta ^{m+1}$ is a standard simplex, in which case the desired result is immediate from the definition.
$\square$