Is any Euclidean ball contained in D some pseudo-hyperbolic ball? We have that the pseudo-hyperbolic distance in the open unit disk $\mathbb D$ is defined by
$$ \rho(z,w) = |\phi_w(z)|,  \qquad \phi_w(z) = \frac{w - z}{1 - \overline w z}$$
where $z,w \in \mathbb D.$ And the pseudo-hyperbolic ball is defined by
$$ \Delta(p,r) = \{z \in \mathbb{D} | \rho(p,z)<r\}, \quad 0<r<1.$$
There is a proposition which says that

for any $p\in\mathbb{D}$ and $0<r<1$, the pseudo-hyperbolic ball is a Euclidean ball with center and radius given by
$$P = \frac{1-r^2}{1-r^2|p|^2}p, \quad R = \frac{1-|p|^2}{1-r^2|p|^2}r,$$
i.e. $\Delta(p,r) = B(P,R).$

Composition Operators and Classical Function Theory says that

The pseudo-hyperbolic distance induces the usual Euclidean topology.

and I'd like to prove it. By the proposition mentioned above, it's clear that the topology induced by pseudo-hyperbolic distance is contained in the Euclidean topology.
However, when I want to prove the inverse relation, I have no idea but to prove that "any Euclidean ball contained in $\mathbb{D}$ is also some pseudo-hyperbolic ball", which means that for any given $P \in \mathbb{D}, 0<R<1, B(P,R) \subset \mathbb{D}, \exists p \in \mathbb{D}$ and $0<r<1$, s.t $$P = \frac{1-r^2}{1-r^2|p|^2}p, \quad R = \frac{1-|p|^2}{1-r^2|p|^2}r$$ holds. But I don't know how to prove it.
My questions are:
Is my assumption above right? If not, how can I prove "The pseudo-hyperbolic distance induces the usual Euclidean topology"?
 A: Any Euclidean ball in $\Bbb D$ contains some pseudo-hyperbolic ball with the same Euclidean center
Since you have proved "the topology induced by pseudo-hyperbolic distance (the pseudo-hyperbolic topology) is contained in the Euclidean topology", you need to prove the Euclidean topology is contained in the pseudo-hyperbolic topology.
For that purpose, you do not have to prove "any Euclidean ball contained in $\Bbb D$ is also some pseudo-hyperbolic ball".  All you need to prove is, intuitively, the neighborhood sets of each point in Euclidean topology are at most as fine as those in the pseudo-hyperbolic topology. Specifically, "for any given $P \in \Bbb D$, $0<R<1$, $B(P,R) \subset \Bbb D$, $\exists p \in \Bbb D$ and $0<r<1$, s.t $\Delta(p,r)=B(P,R')$ for some $R'\le R$." The only difference of this version from your version is "$R=\cdots$"
is replaced by "$R\ge\cdots$".
Suppose $P \in \Bbb D$, $0<R<1$, $B(P,R) \subset \Bbb D$.
Solving the equation with the unknown complex variable $p$, $$P = \frac{1-R^2}{1-R^2|p|^2}p$$
we obtain $p=\frac{-(1-R^2)+\sqrt{(1-R^2)^2+4|P|^2R^2}}{2|P|^2R^2}P$.
Let $R'=\frac{1-|p|^2}{1-R^2|p|^2}R$.
It is straightforward to verify that $0\le |p|<1$, $0<R'\le R$, $\Delta(p,R)=B(P,R')$. We are done.
Any Euclidean ball in $\Bbb D$ is a pseudo-hyperbolic ball 
On the other hand, the answer to the question in the title is yes. Each Euclidean ball in $\Bbb D$ is in fact a pseudo-hyperbolic ball.
Suppose $B(P,R) \subset \Bbb D$, i.e. $P \in \Bbb D$, $0<R<1-|P|$. If $P=0$, then $\Delta(P, R)=B(P,R)$, we are done. Assume $P\not=0$.
Let $\mathscr P:(0,1)\to\Bbb (0, 1), \mathscr P(x)=\frac{-(1-x^2)+\sqrt{(1-x^2)^2+4|P|^2x^2}}{2|P|x^2}$ and
$\mathscr R:(0,1)\to(0,1), \mathscr R(x)=\frac{1-\mathscr P(x)^2}{1-x^2\mathscr P(x)^2}x$.
Verify

*

*$\mathscr P(x)$ is well-defined, i.e. $0<\mathscr P(x)<1$.

*$\mathscr R(x)$ is well-defined, i.e. $0<\mathscr R(x)<1$.

*$\Delta(\mathscr P(x)\frac P{|P|}, x)=(P, \mathscr R(x))$.

Since $\mathscr R(0_+)=0<R<1-|P|=\mathscr R(1_{-})$, there exists $0<r<1$ such that $\mathscr R(r)=R$. Hence $\Delta(p,r)=B(P,R)$, where $p=\mathscr P(x)\frac P{|P|}$.
In fact, one-to-one correspondence
The map $\Delta(p,r)\mapsto B(P,R)$ is a bijection between the pseudo-hyperbolic balls and the Euclidean balls in $\Bbb D$.
It is easy to verify the map is well defined, i.e., $B(P,R)$ is inside $\Bbb D$.
We have proved the map is surjective. The map is, of course, injective since $\Delta(p,r)= B(P,R)$.
