Let singular values of matrix A are written in vector $s = (s_1, \ldots, s_n)$ in decreasing order ($n \ge 2$). Let $x,y$ are orthogonal vectors with $\|x\| = 1$ and $\|y\| = 1$.
What is the biggest value of $\frac{\|Ax- Ay\|}{\|x-y\|}$?
I know about singular values only from wikipedia page, so I need help.
Maybe we can use spectral theory and then apply min-max theorem $s_1 = \max \lambda_i$. I thought like that:
$$x = \sum x_i e_i, \quad y = \sum y_i e_i$$
where $\{e_i\}_1^n$ is a basis from eigenvectors.
Then $$Ax - Ay = A(x-y) = A \sum (x_i-y_i) e_i = \sum (x_i-y_i) Ae_i = \sum \lambda_i(x_i-y_i)e_i$$
$$\|Ax-Ay\| = \|\sum \lambda_i(x_i-y_i)e_i\| \le \sum |\lambda_i| |x_i-y_i| \le s_1\sum|x_i-y_i| $$
$$\|x-y\|^2 = \|x\|^2 + \|y\|^2 = 2 \Rightarrow \|x-y\| = \sqrt2$$
$$\frac{\|Ax-Ay\|}{\|x-y\|} \le s_1 \frac{\sum|x_i-y_i|}{\sqrt2}$$
Is it a good way? And if yes, how to end my solution?
Assuming that you are using Euclidean norm, the largest possible value is $s_1$.
Since $\{x,y\}$ is orthonormal, $x-y=\sqrt{2}u$ for some unit vector $u$. It follows that $$ \frac{\|Ax-Ay\|}{\|x-y\|}=\frac{\|\sqrt{2}Au\|}{\|\sqrt{2}u\|}=\frac{\|Au\|}{\|u\|}\le\|A\|=s_1 $$ and equality holds when $u$ is a singular vector corresponding to the singular value $s_1$, $x=\sqrt{2}(v+u)$ and $y=\sqrt{2}(v-u)$ for some unit vector $v\perp u$.
Remark. Your proof attempt is incorrect. $A$ is not necessarily diagonalisable in the first place. Even if it is diagonalisable, you seem to think that $|\lambda_i|\le s_i$, but this does not survive an obvious sanity check: in case $A$ is nonsingular, since $\prod_i\lambda_i=\det A=\prod_i s_i$, if $|\lambda_i|\le s_i$ for all $i$, we must have $|\lambda_i|=s_i$ for each $i$. But this implies that $\|A\|_F^2=\sum_i|\lambda_i|^2$, which is true only when $A$ is a normal matrix.