What is the name of the following 1st order high-pass IIR filter?
This is a first-order discrete-time Butterworth filter, which can be obtained via the bilinear transform from the following analog prototype filter (with normalized cut-off frequency):

$$H(s)=\frac{s}{1+s}\tag{1}$$

The bilinear transform replaces $s$ by $k\frac{z-1}{z+1}$, where $k$ is chosen such that the discrete-time filter has the specified cut-off frequency $f_c$, resulting in

$$k=\frac{1}{\tan(\pi f_c T)}\tag{2}$$

where $T=1/f_s$ is the sampling interval.

So if you replace $s$ in $(1)$ by

$$s=\frac{1}{\tan(\pi f_c T)}\frac{z-1}{z+1}\tag{3}$$

you obtain the corresponding discrete-time filter with the coefficients given in your question.

EDIT:

OK, here is how you obtain your final coefficients. Let $a=\tan(\pi f_c T)$. Plugging $(3)$ into $(1)$ then gives for the transfer function of the discrete-time filter

$$\begin{align}G(z)=\frac{\frac{1}{a}\frac{z-1}{z+1}}{1+\frac{1}{a}\frac{z-1}{z+1}}&amp;=\frac{z-1}{a(z+1)+z-1}\\&amp;=\frac{z-1}{(1+a)z-(1-a)}\\&amp;=\frac{1}{1+a}\frac{z-1}{z-\frac{1-a}{1+a}}\end{align}\tag{4}$$

which is exactly in the given form with $c_1=1/(1+a)$ and $c_2=(1-a)/(1+a)$.