Why is an inductive limit of bornological spaces bornological? Let $(E_\alpha,\tau_\alpha,g_\alpha)$ be a family of bornological (locally convex) topological vector spaces $(E_\alpha,\tau_\alpha)$, where a LCTVS $E$ is said to be bornological if every circled, bounded subset $A\subset E$ that absorbs every bounded set is a neighborhood of zero (i.e. has nonempty interior), and where $g_\alpha:E_\alpha\to E$ induces an inductive LCTVS structure on $(E,\tau)$.
Then in Shaefer's book Topological Vector Spaces, section II.8.2, we are told that $E$ must be bornological. However, I am slightly confused about a detail of the proof. 

Let $A$ be a convex, circled subset of $E$ absorbing all bounded sets. If $B_\alpha$ is bounded in $(E_\alpha,\tau_\alpha)$ then $g_\alpha(B_\alpha)$ is bounded in $E$, then $A$ absorbs $g_\alpha(B_\alpha)$, whence $g_\alpha^{-1}(A)$ absorbs $B_\alpha$, and so $g_\alpha^{-1}(A)$ contains a neighborhood of $0$ in $E_\alpha$. Since this holds for all $\alpha$, then $A$ contains a neighborhood of $0$ in $E$.

However, I'm very confused about the last step — the inductive limit property does not directly give us the existence of opens in $E$, it merely places a constraint on them. So how do we figure out that $A$ contains a neighborhood of $0$ in $E$?
 A: The topology on $E$ is such that it is the strongest (locally convex TVS) topology $\mathcal{T}$ such that for all $\alpha$: $g_\alpha: E_\alpha \to E$ is continuous. 
This implies that if $O \subseteq E$ is such that that $(g_\alpha)^{-1}[O]$ is open for all $\alpha$ then $O \in \mathcal{T}$. (this holds for convex circled sets, like $A$ per the comment by Jochen) For if this were not the case, we could add it to the topology of $E$ and generate a strictly larger LCTVS topology that still makes all $g_\alpha$ continuous. This explains the last sentence. Also see these notes for some more info, though Schaefer probably has more.
This is analogous to quotient maps (another example of a so-called final topology):  $q: X \to Y$ is a quotient map (between topoloigcal spaces) iff $Y$ has the strongest topology that makes $q$ continuous iff $\forall O \subseteq Y: O \in \mathcal{T}_Y \Leftrightarrow  q^{-1}[O] \in \mathcal{T}_X$, i.e. any set that doesn't contradict the continuity is in the topology already.
