Let $(E, \tau)$ be a perfectly normal space and $\sigma(\tau)$ its Borel $\sigma$-algebra. I'm trying to prove below result. Could you have a check on my attempt?
Theorem: $\sigma(\tau)$ is the smallest $\sigma$-algebra on $E$ such that every continuous function whose domain is $E$ is Borel measurable.
Proof: Let $\mathcal A$ be the smallest $\sigma$-algebra on $E$ such that every continuous function whose domain is $E$ is measurable. Let $(F, \tau')$ be a topological space, $\sigma(\tau')$ its Borel $\sigma$-algebra, and $f:E \to F$ continuous. Notice that $f^{-1} [\sigma(\tau')] = \sigma (f^{-1} [\tau']) \subset \sigma (\tau)$, so $f$ is measurable w.r.t. $\sigma (\tau)$. By minimality of $\mathcal A$, we get $\mathcal A \subset \sigma (\tau)$.
Fix $O \in \tau$. Then $O^c := E \setminus O$ is closed in $\tau$. Because $\tau$ is perfectly normal, there is a continuous function $f:E \to [0, 1]$ such that $O^c =f^{-1} (\{0\})$. Notice that $\{0\}$ is a Borel measurable subset of $[0, 1]$. It follows that $O^c \in \mathcal A$. It follows that $\tau \subset \mathcal A$ and thus $\sigma(\tau) \subset \mathcal A$. This completes the proof.