Since $A$ is an Artin algebra, it has only finitely many indecomposable injective modules in total (up to isomorphism), so there are finitely many in $\Gamma$.

In a locally finite quiver, given any $d\geq0$ and vertex $v$, there are finitely many paths ending at $v$ and having length at most $d$, because there are finitely many choices for each arrow. In particular, $v$ is reachable from only finitely many vertices by such paths.

This means that, for $r$ as in the proof, there only finitely many vertices of $\Gamma$ from which one of the finitely many indecomposable injectives may be reached via a path of length at most $r$. But since the proof shows that every vertex of $\Gamma$ has a path of length at most $r$ to an indecomposable injective, it follows that $\Gamma$ has finitely many vertices.

Since $A$ is an Artin algebra, it has only finitely many indecomposable injective modules in total (up to isomorphism), so there are finitely many in $\Gamma$.

In a locally finite quiver, given any $d\geq0$ and vertex $v$, there are finitely many paths ending at $v$ and having length at most $d$, because there are finitely many choices for each arrow. In particular, $v$ is reachable from only finitely many vertices by such paths.

This means that there only finitely many vertices of $\Gamma$ from which one of the finitely many indecomposable injectives may be reached via a path of length at most $d$, for any fixed $d>0$. Since the proof exhibits an $r$ such that every vertex of $\Gamma$ has a path of length at most $r$ to some indecomposable injective, it follows that $\Gamma$ has finitely many vertices.

Since $A$ is an Artin algebra, it has only finitely many indecomposable injective modules in total (up to isomorphism), so there are finitely many in $\Gamma$.

In a locally finite quiver, given any $d\geq0$ and vertex $v$, there are finitely many paths starting from $v$ and having length at most $d$, because there are finitely many choices for each arrow. In particular, only finitely many vertices are reachable from $v$ by such paths.

This means that, for $r$ as in the proof, there are only finitely many vertices of $\Gamma$ reachable from one of the finitely many indecomposable injectives via a path of length at most $r$. Since the proof shows that every vertex of $\Gamma$ is reachable in this way, it follows that $\Gamma$ has finitely many vertices.

Since $A$ is an Artin algebra, it has only finitely many indecomposable injective modules in total (up to isomorphism), so there are finitely many in $\Gamma$.

In a locally finite quiver, given any $d\geq0$ and vertex $v$, there are finitely many paths ending at $v$ and having length at most $d$, because there are finitely many choices for each arrow. In particular, $v$ is reachable from only finitely many vertices by such paths.

This means that, for $r$ as in the proof, there only finitely many vertices of $\Gamma$ from which one of the finitely many indecomposable injectives may be reached via a path of length at most $r$. But since the proof shows that every vertex of $\Gamma$ has a path of length at most $r$ to an indecomposable injective, it follows that $\Gamma$ has finitely many vertices.