Universal groupoid covering from the trivial subgroup

Apply the subgroup construction to $H=\{1_x\}$. Its objects are individual arrows $a$ from $x$, and an arrow from $a$ to $b$ has the unique possible label $a^{-1}b$. Thus each morphism set is a singleton, in particular every vertex group is trivial. Conversely in any connected groupoid with trivial vertex groups, two arrows $u,v$ with the same endpoints satisfy $u{v}^{-1}=1$, hence $u=v$, and connectedness guarantees existence.

Given a pointed connected cover $(E,e)\to (G,x)$, send object $a$ to the endpoint of its lift from $e$ and arrow $(a,g)$ to the lift of $g$ there. Product lifting makes this a functor over $G$. On each source star both projections are bijections to the same base star, so this functor is a covering; uniqueness follows by lifting from the chosen point.

If $(V,v)\to (G,x)$ is another universal cover in the stated sense, its universal property and that of $U$ give unique pointed functors $F:U\to V$ and $K:V\to U$ over $G$. Both $K\circ F$ and the identity of $U$ are pointed endofunctors over $G$, so uniqueness for $U$ gives $K\circ F={\mathrm{i}\mathrm{d}}_U$. Similarly uniqueness for $V$ gives $F\circ K={\mathrm{i}\mathrm{d}}_V$. Thus $F$ is the unique pointed isomorphism over $G$ from $U$ to $V$; no assertion about the characteristic subgroup of $V$ was assumed.

Put $N=\{1_x\}$. For each $a\in G_x$, $aN{a}^{-1}=\{a{1}_x{a}^{-1}\}=N$, so $N$ is normal. Its characteristic-subgroup realization in $U$ and the normal-deck-quotient lemma therefore make $U$ normal with $G_x\mathrm{/}N\cong \mathrm{D}\mathrm{e}\mathrm{c}\mathrm{k}(U\to G)$. Each coset $aN=\{a\}$ is a singleton, so $\varphi :G_x\mathrm{/}N\to G_x$, $\varphi (aN)=a$, is well defined and bijective. The quotient product gives $\varphi ((aN)(bN))=ab=\varphi (aN)\varphi (bN)$, so $\varphi$ is a group isomorphism. Combining these isomorphisms proves $\mathrm{D}\mathrm{e}\mathrm{c}\mathrm{k}(U\to G)\cong G_x$ with the stated deck convention. For the isomorphism $F$ in step 3.1, $T\mapsto F\circ T\circ K$ identifies the deck groups, preserves ordinary composition and intertwines their actions on the corresponding fibres; hence $V$ is also normal with the same deck-group conclusion.