estimate of a holomorphic function in a unit disc Prove that if $a\not \in \bar{\mathbb{D}}$, then 
$$\inf_{c\in\mathbb{C}}\left(\sup_{z\in\mathbb{D}}\left|\frac{z-c}{z-a}\right|\right)=\frac{1}{|a|}$$
My Observation:
If $a\not \in \bar{\mathbb{D}}$ and $z \in \mathbb{D}$ then $\sup_{z\in\mathbb{D}}\left|\dfrac{z-c}{z-a}\right| = \left|\dfrac{z_0-c}{z_0-a}\right|$ where $z_0 \in \partial \mathbb{D}$ is the point at which the line joining the origin and $a$ meets the unit circle. This is because $f(z)= \frac{z-c}{z-a}$ is holomorphic in $\mathbb{D}$ and hence by Maximum Modulus principle, attains its maximum on the unit circle. This implies that $|z_0-a| = |a| -1$ and hence we need to show that $\inf_{c \in \mathbb{C}}|z_0-c|= \frac{|a|-1}{|a|}=1-\frac{1}{|a|}$. This can happen if $c$ is the point symmetric to $a$ with respect to the unit circle that is $c = \frac{1}{\bar{a}}$. I do not know how to prove that or if there is some better method altogether. 

I am not convinced by your argument in favor of $$\sup_{z\in\mathbb{D}}\left|\dfrac{z-c}{z-a}\right| = \left|\dfrac{z_0-c}{z_0-a}\right|$$
Take $a=3$ and $c=2$; then at $z=1$ we have $\left|\dfrac{z-c}{z-a}\right|=\frac12$ but at $z=-1$ it's $\left|\dfrac{z-c}{z-a}\right|=\frac34$. 

The function $f$ is a fractional linear transformation. Since its domain is a disk, it is natural to relate it somehow to automorphisms of a disk, which have the form (up to a constant factor) 
$$z\mapsto \frac{z-\beta}{1-\bar \beta z}$$
A step toward this form: multiply both sides of the statement by $|a|$ to bring it to 
$$\inf_{c\in\mathbb{C}}\left(\sup_{z\in\mathbb{D}}\left|\frac{z-c}{1-z/a}\right|\right)=1 \tag{1}$$
Clearly, we are dealing with $\beta=1/\bar a$ here. Write
$$
\phi(z) = \frac{z-1/\bar a}{1-z/a}
$$
(an automorphism of $\mathbb D$) and record that 
$$ \left|\frac{z-c}{1-z/a}\right|=
\left| \phi(z) - \frac{c-1/\bar a}{1-z/a}\right|
 \tag{2}$$
If $c=1/\bar a$, the supremum of (2) over $z\in\mathbb D$ is  $1$. Suppose there is $c\in\mathbb C$ for which this supremum is less than $1$. Then for $z\in\partial \mathbb D$ we have 
$$
\left|\phi(z) - \frac{c-1/\bar a }{1-z/a}\right| < 
\left| \phi(z)  \right|
$$
By Rouche's theorem, the function $\frac{c-1/\bar a }{1-z/a}$ has as many zeros in $\mathbb D$ as $\phi$ does, namely one. This is absurd.
