First, suppose $\mathcal{F}$ is not reflexive, say, $(w,w)\notin R$. Let $M$ be a model based on $\mathcal{F}$ such that $V(p)=\{u\mid wRu\}$, where $p$ is a propositional variable. By the construction of $V(p)$, we see that for all $u$ such that $wRu$, we have $\models_{u}p$, so $\models_{w}\square p$. But since $w\notin V(p)$, $\not\models_{w}p$. This means that $\not\models_{w}\square p\to p$.

Conversely, let $\mathcal{F}$ be a reflexive frame, and $M$ any model based on $\mathcal{F}$, with $w$ a world in $M$. Suppose $\models_{w}\square A$. Then for all $u$ such that $wRu$, $\models_{u}A$. Since $wRw$, we get $\models_{w}A$. Therefore, $\models_{w}\square A\to A$. ∎

Since any theorem in T is deducible from a finite sequence consisting of tautologies, which are valid in any frame, instances of T, which are valid in reflexive frames by the proposition above, and applications of modus ponens and necessitation, both of which preserve validity in any frame, whence the result. ∎

We show that the canonical frame $\mathcal{F}_{{\textbf{T}}}$ is reflexive. For any maximally consistent set $w$, if $A\in\Delta_{w}:=\{B\mid\square B\in w\}$, then $\square A\in w$. Since T contains $\square A\to A$, we get that $A\in w$ by modus ponens and the fact that $w$ is closed under modus ponens. Therefore $wR_{{\textbf{T}}}w$, or $R_{{\textbf{T}}}$ is reflexive. ∎