The upward closure of a filter base is the smallest filter containing it

Every member of $\langle \mathcal{B} \rangle$ is a subset of $X$ by definition, so $\langle \mathcal{B} \rangle \subseteq \mathcal{P}(X)$; and $X \in \langle \mathcal{B} \rangle$, since $\mathcal{B}$ has a member $B$ and $B \subseteq X$.

$\emptyset \notin \langle \mathcal{B} \rangle$: if $B \subseteq \emptyset$ for some $B \in \mathcal{B}$ then $B = \emptyset$, and $\emptyset \notin \mathcal{B}$.

$\langle \mathcal{B} \rangle$ is closed upward in $X$: if $A \in \langle \mathcal{B} \rangle$, say $B \subseteq A$ with $B \in \mathcal{B}$, and $A \subseteq A' \subseteq X$, then $B \subseteq A'$, so $A' \in \langle \mathcal{B} \rangle$.

$\langle \mathcal{B} \rangle$ is closed under pairwise intersection: given $A_1, A_2 \in \langle \mathcal{B} \rangle$ pick $B_1, B_2 \in \mathcal{B}$ with $B_i \subseteq A_i$, and then $B_3 \in \mathcal{B}$ with $B_3 \subseteq B_1 \cap B_2 \subseteq A_1 \cap A_2 \subseteq X$, so $A_1 \cap A_2 \in \langle \mathcal{B} \rangle$.

$\mathcal{B} \subseteq \langle \mathcal{B} \rangle$, because $B \subseteq B$ and $B \subseteq X$ for every $B \in \mathcal{B}$.

If $\mathcal{F}$ is a filter on $X$ with $\mathcal{B} \subseteq \mathcal{F}$ and $A \in \langle \mathcal{B} \rangle$, say $B \subseteq A \subseteq X$ with $B \in \mathcal{B} \subseteq \mathcal{F}$, then upward closure of $\mathcal{F}$ gives $A \in \mathcal{F}$; hence $\langle \mathcal{B} \rangle \subseteq \mathcal{F}$.

Every filter $\mathcal{F}$ on $X$ is a filter base: $X \in \mathcal{F}$ gives $\mathcal{F} \neq \emptyset$, properness gives $\emptyset \notin \mathcal{F}$, and for $B_1, B_2 \in \mathcal{F}$ the member $B_3 = B_1 \cap B_2$ of $\mathcal{F}$ satisfies $B_3 \subseteq B_1 \cap B_2$.

$\langle \mathcal{B} \rangle$ satisfies all four filter axioms, so it is a filter on $X$.

For a filter $\mathcal{F}$ on $X$, applying step 1.5 and step 1.6 to the filter base $\mathcal{F}$ gives $\mathcal{F} \subseteq \langle \mathcal{F} \rangle$ and $\langle \mathcal{F} \rangle \subseteq \mathcal{F}$, that is $\langle \mathcal{F} \rangle = \mathcal{F}$.

So $\langle \mathcal{B} \rangle$ is a filter containing $\mathcal{B}$ and contained in every filter that contains $\mathcal{B}$, hence the smallest such filter, and every filter is a filter base generating itself.