Can the details for the proof of Proposition 15.8 of Daniel Bump's Lie Groups, 2nd Edition, be filled in without Cartan's closed-subgroup theorem?

This is Proposition 15.8 and its proof from p.106 of Daniel Bump's Lie Groups, 2nd Edition. $\textbf{Proposition 15.8.}$ Let $G$ be a compact Lie group and $T$ a maximal torus. Then $N(T)$ is a closed subgroup of $G$. The connected component $N(T)^◦$ of the identity in $N(T)$ is $T$ itself. The quotient $N(T)/T$ is a finite group. $\textit{Proof.}$ We have a homomorphism $N(T)\to\operatorname{Aut}(T)$ in which the action is by conjugation. By Proposition 15.7, $\operatorname{Aut(T)}\cong\mathrm{GL}(r,\Bbb Z)$ is discrete, so any connected group of automorphisms must act trivially. Thus, if $n\in N(T)^{\circ}, n$ commutes with $T$ . If $N(T)^{\circ}\neq T$, then it contains a one-parameter subgroup $\Bbb R\ni t\to n(t)$, and the closure of the group generated by $T$ and $n(t)$ is a closed commutative subgroup strictly larger than $T$. By Theorem 15.2, it is a torus, contradicting the maximality of $T$. It follows that $T=N(T)^{\circ}$. The quotient group $N(T)/T$ is both discrete and compact and hence finite. I do not know how to justify the claim that if $N(T)^{\circ}\neq T$, then it contains a one-parameter subgroup $\Bbb R\ni t\to n(t)$ such that the closure of the group generated by $T$ and $n(t)$ is a closed commutative subgroup strictly larger than $T$ without invoking Cartan's closed-subgroup theorem. Here is my attempt: We first prove the claim that if $G$ is a connected Lie group that contains the Lie subgroup $H$ such that their dimensions are equal, then $H=G$. Since their dimensions are equal, the differential of the inclusion is invertible everywhere. By the Inverse Function Theorem, the inclusion is a local diffeomorphism. Thus $H$ is an open subgroup of $G$. It is well known that every open subgroup is clopen. Hence $H$ is a clopen subgroup containing the identity. This implies $H$ contains the connected component of the identity which is precisely $G$ as $G$ is connected. Then we prove that $N(T)$ is a closed subgroup. Pick a generator $t$ of $T$. The statement follows if we can prove that $N(T)$ is the inverse image of $T$ under the continuous map $g\to gtg^{−1}$. Clearly, if $g\in N(T)$, then $g$ is in the inverse image. Conversely, assume $g$ is in the inverse image. By closure of group operations and closure, the torus $T$ contains $gTg^{-1}$. Since $T$ is connected and conjugation is a diffeomorphism which preserves dimension, by our claim in the previous paragraph, we have $gTg^{-1}=T$. Hence $g\in N(T)$. By Cartan's closed-subgroup theorem, the closed subgroup $N(T)$ is a Lie subgroup of $G$. Now we argue that $N(T)$ is a Lie group. It is well-known that the connected component of a topological group is a subgroup of the group. In particular, $N(T)^{\circ}$ is a subgroup of $N(T)$. Since $N(T)$ is a Lie group, it is locally path-connected. Thus the connected component $N(T)^{\circ}$ is open in $N(T)$ and thereby a Lie subgroup of $N(T)$. Finally, we construct a one-parameter subgroup $\Bbb R\ni t\to n(t)$ contained in $N(T)^{\circ}$ such that the closure of the group generated by $T$ and $n(t)$ is a closed commutative subgroup strictly larger than $T$. Since $N(T)^{\circ}$ is connected, the strict inclusion $T\subsetneq N(T)^{\circ}$ induces the strict inclusion of lie algebras $\mathfrak{t}\subsetneq\mathfrak{n}$. Otherwise, our claim in the first paragraph would imply $T=N(T)^{\circ}$. Pick some $X\in\mathfrak{n}\setminus\mathfrak{t}$. Define $n(t)=\exp(tX)$ for $t\in\Bbb R$. Since $X\notin\mathfrak{t}$, we have $n(t)\notin T$. It is straightforward to verify that the group generated by $T$ and $n(t)$ is abelian. By continuity of group operations, it follows that the closure of the group is a closed abelian subgroup. Moreover, it is strictly larger than $T$ since it contains $n(t)$. I was wondering if anyone knows how to justify the claim without Cartan's closed-subgroup theorem. The reason as to why I believe it might be possible is because the author has mentioned this theorem earlier in $\textit{remark 7.2.}$ on p.45 and only proves of an abelian subgroup in Theorem 15.2 on p.105-106, just before proving Proposition 15.8, which seems to suggest we do not need the closed-subgroup theorem for general groups. It should be noted that $N(T)^{\circ}$ need not be abelian (consider, for example, $G=\mathrm{SU}(2)$), so we could not invoke the abelian case of the closed-subgroup theorem to claim that $N(T)$ is a Lie subgroup.