I don't understand the argument at 26:00-27:14. Where did he use the definition of the standard $\mathbb{R}^d$ topology? It seems that the empty set is in any family of sets by that proof which is clearly nonsense.
A youtube snapshot is helpful - compare the definition of $\mathcal U$ on the left side of the board to the right side where he proves $\emptyset \in \mathcal U$.
On the left hand side of the board, the standard topology on $\mathbb{R}^d$ is defined by: A subset $U \subseteq \mathbb{R}^d$ is in the standard topology, if and only if $\forall p \in U: \exists r > 0$ s.t. $B_r(p) \subseteq U$.
With this definition, the empty set is indeed in the standard topology. We just need to check that $\forall p \in \emptyset: \exists r > 0$ s.t. $B_r(p) \subseteq \emptyset$. However, this is trivially true, since there are no $p \in \emptyset$.
More precisely, the statement is true since we have
$$\forall p,\exists r > 0: p \in \emptyset \Rightarrow B_r(p) \subseteq \emptyset$$
However, this means that for every $p$ we have $\text{false} \Rightarrow \text{something}$, which by Ex falso quodlibet, is a true sentence.
Also, the proof does not show that the empty set is in any family of sets. This is not true. For example, consider the following collection of subsets of $\mathbb{N}$ defined via $S:= \{U \subseteq \mathbb{N} \vert 1 \in U\}$. It does not contain the empty set.
However, the proof does show that for any collection $A$ of sets defined by $U \in A$ if and only if $\forall p \in U: \text{something}$ the empty set is a member.