An attempt of proof, please review carefully and tell me your opinion. Thanks and kind regards.
Recall that the $n$-torus $T^n$ is defined as the quotient $\Bbb{R}^n/G$ where $G$ is the group of integer translations in $\Bbb{R}^n$. It can be identified with the product of $n$ circles:
$$
T^n = \underbrace{S^1 \times \ldots \times S^n}_{n \text{ times}}.
$$
We can then define a natural projection $\pi: \Bbb{R}^n \longrightarrow T^n$ given by
$$
\pi(x) = (e^{ix_1}, \ldots, e^{ix_n}), \quad x = (x_1, \ldots, x_n) \in \Bbb{R}^n.
$$
For $u = (u_1, \ldots, u_n) \in T_x\Bbb{R}^n = \Bbb{R}^n$ we have
$$
d\pi_x(u) = J(x)u = i(u_1 e^{i x_1}, \ldots, u_n e^{i x_n}).
$$
where $J(x)$ is the Jacobian matrix of $\pi$ at $x$ given by
$$
J(x) = \begin{bmatrix}
ie^{ix_1} & 0 & \cdots & 0 \\
0 & ie^{i x_2} & \cdots & 0 \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \cdots & i e^{i x_n}
\end{bmatrix}.
$$
We can just define
$$
\langle d\pi_x(u), d \pi_x(v) \rangle_{\pi(x)} = \langle u, v \rangle, \quad u, v \in T_x\Bbb{R}^n = \Bbb{R}^n
$$
where $\langle \cdot, \cdot \rangle$ denotes the inner product of the Euclidean space. To get a local isometry, it suffices to restrict $\pi$ to a neighborhood of $x$ such that it is a diffeomorphism.
We now show that the identity map $i: \Bbb{R}^n/G \longrightarrow T^n$ is an isometry, that is, we show that the two metrics are the same. Let $u,v \in T_pT^n$. Then
\begin{align*}
\langle u, v \rangle_{(e^{i x_1}, \ldots, e^{i x_n})} & = \sum_1^n \langle d \pi_j (u), d \pi_j(v) \rangle_{e^{i x_j}} \\
& = \sum_1^n \langle u_j e^{i (x_j + \pi/2)}, v_j e^{i (x_j + \pi/2)} \rangle_{e^{i x_j}} \\
& = \sum_1^n u_j v_j \\
& = \langle u, v \rangle,
\end{align*}
which completes the proof.
Best Answer
You are defining $\mathbb{P}^n = \mathbb{S}^n / p\sim A(p)$. Let's assume that you have verified to your satisfaction that this action is "nice" enough to produce a smooth quotient manifold.
Let's consider any point $p\in\mathbb{P}^n$. The only thing we know about $p$ is that it's an equivalence class $p = [x] = \{x_1, x_2\}$ which are related by $x_i = A(x_j)$. The tangent spaces $T_{x_1}\mathbb{S}^n$ and $T_{x_2}\mathbb{S}^n$ are related by $dA$, and via this identification we can consider either of them a model for $T_p\mathbb{P}^n$. So if we want to put a metric on $T_p\mathbb{P}^n$, there's really only one thing we can do: pick the metric on one of the tangent spaces $T_{x_i}\mathbb{S}^n$.
Now we worry! We've made a choice of identification. Does it agree with all the other possible choices we could have made? In fact, yes, precisely because $A$ is an isometry. The metric on $T_{x_j}\mathbb{S}^n$ pulled back by $dA$ is exactly what we would have picked if we had chosen $T_{x_j}\mathbb{S}^n$ to define our metric.
This strategy works whenever $Y = X/\Gamma$, where $X$ is a Riemannian manifold and $\Gamma$ is a discrete group acting freely on $X$ by isometries. Then the quotient map is a covering and the covering neighborhoods are all isometric, so the choice of metric is natural.
The general case $Y = X/G$, $X$ Riemannian and $G$ is any group acting by isometries, is more complicated.