By definition
\begin{equation}
\text{int} \langle E \rangle
: = \bigcup \{ U \in \mathcal{T} : U \subseteq E \}
\end{equation}
Since every set in the union $\text{int} \langle E \rangle$ is required to be contained in $E$, then result
R4152: implies the inclusion $\text{int} \langle E \rangle \subseteq E$, as claimed. $\square$
