Proof. Let $\mathscr{T}$ be the collection of all topologies satisfying (1), then $\mathcal{T}\in \mathscr{T}$, so $\mathscr{T}\ne \emptyset$. Let $\mathcal{T}_{B}$ be the projective topology induced by $\mathscr{T}$.

(1): Since $\mathcal{T}_{B} \supset \mathcal{T}$, $\mathfrak{B}(E, \mathcal{T}) \supset \mathfrak{B}(E, \mathcal{T}_{B})$. Let $B \in (E, \mathcal{T})$ and $U \in \cn_{\mathcal{T}_B}(0)$, then there exists $\mathscr{S}\subset \mathscr{T}$ finite and $\bracsn{U_{\mathcal{S}}}_{\mathcal{S} \in \mathscr{S}}$ such that $\bigcap_{\mathcal{S} \in \mathscr{S}}U_{\mathcal{S}}\subset U$. In which case, since $\mathfrak{B}(E, \mathcal{T}) = \mathfrak{B}(E, \mathcal{S})$ for all $\mathcal{S}\in \mathscr{S}$, $\bigcap_{\mathcal{S} \in \mathscr{S}}U_{\mathcal{S}}$ absorbs $B$, so $U$ absorbs $B$ as well.

(U): By (U) of the projective topology.

(3): Let $\rho: E \to [0, \infty)$ be a seminorm that is bounded on bounded sets, then the topology induced by $\rho$ satisfies (1). By (U), $\rho$ is continuous with respect to $\mathcal{T}_{B}$, so $(E, \mathcal{T}_{B})$ is a bornological space.

(4): Let $\mathcal{S}$ be the inductive topology on $E$ induced by $\bracsn{\iota_B}_{B \in \mathcal{B}}$. Let $B \in \mathfrak{B}(E, \mathcal{T})$, $U \in \cn_{\mathcal{S}}(0)$, and $\rho: E \to [0, \infty)$ be the gauge of $U$. Since $\iota_{\ol B}\in L(E_{\ol B}; E, \mathcal{S})$, there exists $\lambda > 0$ such that $\rho \le \lambda \iota_{\ol B}\circ \rho_{\ol B}$. Thus $2\lambda U \supset B$ and $(E, \mathcal{S})$ satisfies (1). By (U), $\mathcal{T}_{B} \supset \mathcal{S}$.

On the other hand, since $\mathfrak{B}(E, \mathcal{T}_{B}) = \mathfrak{B}(E, \mathcal{T})$, for each $B \in \mathcal{B}$, $\iota_{B}: E_{B} \to (E, \mathcal{T}_{B})$ is continuous. Therefore by (U) of the inductive topology, $\mathcal{S}\supset \mathcal{T}_{B}$.$\square$