Let $f: [0,1] \to \mathbb R$ be a continuously differentiable function. For $k, n \in \mathbb N$, we can define the set $S_{k,n}$ as:
$\{ x \in [0,1]: $ for any $y \in [x - \frac{1}{k}, x + \frac{1}{k}] \cap [0,1]$ it holds $|f(x) - f(y) | \leq \frac{1}{n}|x-y|\}$
How can I show that $S_{k,n} \subset S_{k+1, n}$ and that $S_{k, n+1} \subset S_{k,n}$ for every $k,n \in \mathbb N$?
Write down the definition of the two sets, in particular, what it means for $x$ to be in them. You prove that $A\subseteq B$ by showing that if $x\in A$, then $x\in B$.
For example:
Let's prove $S_{k,n}\subseteq S_{k+1,n}$
What we know:
If $x\in S_{k,n}$, then we know that for all $y \in \left[x - \frac{1}{k}, x + \frac{1}{k}\right]$, we have $|f(x) - f(y) | \leq \frac{1}{n}|x-y|$
What we need to prove:
We need to prove that $x\in S_{k+1, n}$. This is true if and only if, for all $y\in\left[x-\frac{1}{k+1}, x+\frac1{k+1}\right]$, we have $|f(x)-f(y)|\leq\frac1n|x-y|$.
As usual, when you need to prove something for all $y$, your proof should start with:
and conclude with
Your next step is to try and construct this proof. Good luck, and if you get stuck, let us know!