A vertex $v$ is extendible if and only if $G − v$ is a forest. I need help understanding the solution to this problem. This problem has been answered here, however, my doubt is not addressed.
Problem: Let $G$ be a connected Eulerian  graph with at least $3$ vertices. A vertex $′v′$ in $G$ is extendible if every trail beginning at $′v′$ can be extended to form an Eulerian Circuit.
Prove following statement: A vertex $v\in V(G)$ is extendible if  and only if $G-v$ is a forest.
Solution :
Necessity: We prove the contrapositive. If $G − v$ is not a forest, then
$G − v$ has a cycle $C$ . In $G − E(C)$ , every vertex has even degree, so the
component of $G − E(C)$ containing $v$ has an Eulerian circuit. This circuit
starts and ends at $v$ and exhausts all edges of $G$ incident to $v$, so it cannot
be extended to reach $C$ and complete an Eulerian circuit of $G$.
Sufficiency: If $G −v$ is a forest, then every cycle of $G$ contains $v$ . Given a
trail $T$ starting at $v$, extend it arbitrarily at the end until it can be extended
no farther. Because every vertex has even degree, the process can end only
at $v$. The resulting closed trail $T'$ must use every edge incident to $v$, else it
could extend farther. Since $T'$ is closed, every vertex in $G − E(T' )$ has even
degree. If $G − E(T)$ has any edges, then minimum degree at least two in a
component of $G − E(T)$ yields a cycle in $G − E(T')$; this cycle avoids $v$, since
$T'$ exhausted the edges incident to $v$. Since we have assumed that $G − v$
has no cycles, we conclude that $G − E(T')$ has no edges, so $T'$ is an Eulerian
circuit that extends $T$.
Please explain the necessity part, especially the highlighted part.
 A: The definition of extendable says that every trail starting at $v$ can be extended to an Eulerian circuit of $G$. But the Eulerian circuit we find in $G - E(C)$ is a trail starting at $v$ which cannot be extended any further, contra the definition.
In more detail: removing the cycle $C$ only changes degrees by $2$, therefore, by the familiar necessary and sufficient condition for the existence of an Euler circuit, as $G$ was Eulerian, and so had all vertices of even degree, $G - E(C)$ has all vertices of even degree and so each connected component of $G - E(C)$ is Eulerian. This provides an Euler circuit $R$ in the component of $v$ in $G - E(C)$ - that is, a closed trail which passes through every edge. In particular, we may choose this trail to start (and so end) at $v$; and in particular it passes through every edge incident to $v$ (note that this includes every edge incident to $v$ in the original graph $G$, as the cycle $C$ that we removed lies in $G - v$).
Then, in $G$, we may start at $v$ and follow $R$ round back to $v$ again. We have used up every edge incident to $v$, but we have have not visited any edge of $C$. So the trail $R$ cannot be extended to an Euler circuit of $G$.
