I am stuck with the following problem. I am given the $d$-dimensional real projective space $\mathbb{R}P^d$ as the set of equivalence classes of lines in $\mathbb{R}^{d+1}$, i.e. \begin{equation} \mathbb{R}P^d=\mathbb{R}^{d+1}/{\sim} \end{equation} with $[x]=[x']\iff x=ax'$ for some $a\neq 0$. We equip now $\mathbb{R}P^d$ with two different topologies. For one, the quotient topology $\mathcal{Q}$ which it inherits from the Euclidean space $\mathbb{R}^d$ and, on the other hand with the natural topology as a manifold. I mean by this the following. Given the sets $U_i:=\{[x]\in \mathbb{R}P^d: x_i\neq0\}$ and the charts. \begin{align} \kappa_i: \enspace& U_i \rightarrow \mathbb{R}^{d} \\ &[x_0,\dots, x_d]\mapsto \frac{1}{x_i}(x_0,\dots, x_d) \end{align} where we just omit the $(i+1)$st entry that is $1=\frac{x_i}{x_i}$. We define a topology $\mathcal{T}$ on $\mathbb{R}P^d$ as the topology that makes the $\kappa_i$ for $i=0,\dots,d$ to homomorphisms, i.e. \begin{equation} \mathcal{T} = \{O\subset \mathbb{R}P^d: \kappa_i(U_i\cap O) \enspace\text{ open in} \enspace\mathbb{R}^d \enspace\text{for all }i=0,\dots,d\} \end{equation}
I have already shown that if a topology makes the projective space locally Euclidean, it is the same as $\mathcal{T}$.
My problem now is: Are these two topologies the same?
So far, I have not managed to show any direction. Maybe the following could work for $\mathcal{T}\subset \mathcal{Q}$. I know that the quotient topology is the finest topology for which the projection map $\pi: \mathbb{R}^{d+1}\to \mathbb{R}P^d$ is continuous. Suppose now that $O\in \mathcal{T}$, then $O=\kappa_i^{-1}(O')$ some some $i=0,\dots, d$ and $O'$ open in $\mathbb{R}^d$ because homeomorphisms are open maps. Thus, \begin{equation} \pi^{-1}(O)= \{ \lambda x : x\in O, \lambda\neq 0 \}=\{ \lambda \kappa^{-1}(x): \lambda\neq 0, x\in O' \}=\bigcup_{\lambda\neq 0} \{\lambda O'\times\{x_i=1\}\} \end{equation} where the last equal sign just means, that the open set $O'$ is now seen as a set in $\mathbb{R}^{d+1}$ with $i$th component $1$ (and then modulated by $\lambda$ and this should intuitively be open as $\lambda$ runs over all non-zero reals and $O'$ has been formerly open.
Is that correct? For the other direction, I have tried to look at some $O\in \mathcal{Q}$ meaning that $O=\{[x]: x\in O'\}$ and $O'$ is open in $\mathbb{R}^{d+1}$. Then I look for some $i=0,\dots, d$ at \begin{equation} \kappa_i(O\cap U_i) = \bigcup_{\substack{x\in O \\ x_i\neq 0}} \Bigl( \frac{x_0}{x_i}, \dots, \frac{x_d}{x_i} \Bigr) \end{equation} which is somehow a scaled version of $O\cap \{x_i\neq 0\}$, where the $i$th coordinate has been left out. I have no idea why this should or should not be open.
I would appreciate any help. Thank you in advance
For context: The original goal is to show that the projective space is a compact manifold, however, in the exercise the topology is not (as usually) given with the quotient topology but this other one I defined. Thus, if a can show, the topologies are the I could show that the projective space equipped with the