Proof.
We first show that, if $f: X \rightarrow Y$ is a morphism of $\operatorname{\mathcal{E}}$ satisfying conditions $(1)$ and $(2)$, then $f$ exhibits $Y$ as a $\operatorname{\mathcal{E}}'$-reflection of $X$. It follows from condition $(2)$ that $U(Y)$ belongs to $\operatorname{\mathcal{C}}'$, so that $Y$ belongs to $\operatorname{\mathcal{E}}'$. It will therefore suffice to show that for each object $Z \in \operatorname{\mathcal{E}}$, precomposition with $f$ induces a homotopy equivalence $\theta : \operatorname{Hom}_{\operatorname{\mathcal{E}}}(Y,Z) \rightarrow \operatorname{Hom}_{\operatorname{\mathcal{E}}}(X,Z)$. Let us abuse notation by identifying $\theta $ with the restriction map $\{ f\} \times _{ \operatorname{Hom}_{\operatorname{\mathcal{E}}}(X,Y) } \operatorname{Hom}_{\operatorname{\mathcal{E}}}(X,Y,Z) \rightarrow \operatorname{Hom}_{\operatorname{\mathcal{E}}}(X,Z)$, so that we have a commutative diagram of Kan complexes
\[ \xymatrix@R =50pt@C=50pt{ \{ f\} \times _{ \operatorname{Hom}_{\operatorname{\mathcal{E}}}(X,Y) } \operatorname{Hom}_{\operatorname{\mathcal{E}}}(X,Y,Z) \ar [r]^-{ \theta } \ar [d] & \operatorname{Hom}_{\operatorname{\mathcal{E}}}(X,Z) \ar [d] \\ \{ U(f) \} \times _{ \operatorname{Hom}_{\operatorname{\mathcal{C}}}( U(X), U(Y) ) } \operatorname{Hom}_{\operatorname{\mathcal{C}}}( U(X), U(Y), U(Z) ) \ar [r]^-{ \overline{\theta } } & \operatorname{Hom}_{\operatorname{\mathcal{C}}}( U(X), U(Z) ). } \]
Assumption $(1)$ guarantees that this diagram is a homotopy pullback square (Proposition 5.1.2.1), and assumption $(2)$ guarantees that $\overline{\theta }$ is a homotopy equivalence of Kan complexes. Applying Corollary 3.4.1.5, we conclude that $\theta $ is also a homotopy equivalence.
We now show that $\operatorname{\mathcal{E}}'$ is a reflective subcategory of $\operatorname{\mathcal{E}}$. Fix an object $X \in \operatorname{\mathcal{E}}$. Since $\operatorname{\mathcal{C}}'$ is a reflective subcategory of $\operatorname{\mathcal{C}}$, there exists a morphism $\overline{f}: U(X) \rightarrow \overline{Y}$ in $\operatorname{\mathcal{C}}$ which exhibits $\overline{Y}$ as a $\operatorname{\mathcal{C}}'$-reflection of $U(X)$. Since $U$ is a cocartesian fibration, we can write $\overline{f} = U(f)$ for some $U$-cocartesian morphism $f: X \rightarrow Y$ of $\operatorname{\mathcal{E}}$. By construction, the morphism $f$ satisfies conditions $(1)$ and $(2)$, and therefore exhibits $Y$ as an $\operatorname{\mathcal{E}}'$-reflection of $X$.
To complete the proof, it will suffice to show that if $h: X \rightarrow Z$ is another morphism which exhibits $Z$ as a $\operatorname{\mathcal{E}}'$-reflection of $X$, then $h$ also satisfies conditions $(1)$ and $(2)$. By virtue of Remark 6.2.2.3, there exists a $2$-simplex
\[ \xymatrix@R =50pt@C=50pt{ & Y \ar [dr]^{g} & \\ X \ar [ur]^{f} \ar [rr]^{h} & & Z, } \]
of $\operatorname{\mathcal{E}}$, where $g: Y \rightarrow Z$ is an isomorphism of $\operatorname{\mathcal{E}}'$. In particular, $g$ is $U$-cocartesian (Proposition 5.1.1.9), so that $h$ satisfies $(1)$ by virtue of Corollary 5.1.2.4. Since $U(g)$ is an isomorphism in $\operatorname{\mathcal{C}}'$, condition $(2)$ follows from Remark 6.2.2.3.
$\square$