I am reading part of the proof that a Brownian motion is nowhere differentiable for almost all $\omega \in \Omega$ by Rene Schilling.
Assume that the function $w$ is differentiable at $t_0\in [0,n)$. Then $$\exists \delta>0 \; \exists L>0 \; \forall t \in B(t_0,\delta) \; : |w(t)-w(t_0)|\le L|t-t_0|. $$ Consider for sufficiently large values of $k\ge 1$ the grid $\{j/k:j=1,\dots,nk\}$. Then there exists a smallest index $j=j(k) $ such that $$t_0 \le j/k \; \text{and }j/k , \dots (j+3)/k \in B(t_0,\delta).$$
For $i=j+1,j+2,j+3$ we get therefore $$|w(i/k)-w((i-1)/k)|\le |w(i/k)-w(t_0)|+|w(t_0)-w((i-1)/k)|\le L(|i/k-t_0|+|(i-1)/k-t_0|\le L(4/k+3/k)=7L/4.$$
I can't figure out the last part, why are $|i/k-t_0|\le 4/k$ and $|(i-1)/k-t_0|\le 3/k$ from what we have?
By assumption of "smallest index $j=j(k)$" we obtain $$\frac{j-1}{k}< t_{0}\leq \frac{j}{k}$$ and so $$\left|\frac{i}{k}-t_{0}\right|\leq\left|\frac{j+3}{k}-\frac{j-1}{k}\right|=\frac{4}{k}.$$ Similar calculations occurs for the other term.