In the proof of the quoted Proposition, we have a connected graph $X$ and a sequence of subgraphs $X_0 \subset X_1 \subset \cdots$ such that $\cup_i X_i$ is both open and closed. Then Hatcher deduces that $X =\cup_i X_i$.
Question: Why do we need both the openess and closedness of $\cup_i X_i$ in order to deduce the equality $X =\cup_i X_i$? It seems to me that only openess is enough, since if $X \neq \cup_i X_i$, then by connectedness of $X$ there is an edge $e$ of $X$, not contained in $\cup_i X_i$ whose vertex is a vertex of $\cup_i X_i$ and by openess of $\cup_i X_i$ some points of $e$ must be contained in $\cup_i X_i$ and so the entire edge must be contained in $\cup_i X_i$; contradiction.
You are saying $\cup_i X_i$ is a subgraph, therefore the entire edge must be contained in $\cup_i X_i$. But a subset is a subgraph if and only if it is closed. These are the really the same thoughts.
Hatcher probably chose his wording because it is a general (and to most people, quite familiar) topological fact that the only nonempty subspace of a connected space which is open and closed is the entire space itself.