Let $\{(a_i, b_i)\}_{i=1}^n \subseteq [0, 1]\times [0, 1]$ be given such that $d(a_i, b_i)<\delta$, $a_i\neq b_i$, for all $1\leq i\leq n$.
Is there a homeomorphism $h:[0, 1]\to [0, 1]$ such that for all $x\in [0, 1]$, $d(h(x), x)<\delta$ and $h(a_i)=b_i$ for all $1\leq i\leq n$.
Please help me to know it.
In the square $[0,1]\times[0,1]$, draw the lines $y = x+\delta$ and $y = x-\delta$. We want all our points $(a_i, b_i)$, as well as the graph of $h$, to lie between these lines. We also want the graph of $h$ to go through all our points.
Clearly, if $\delta<1$, then $h$ must be strictly monotonically increasing in order to be a homeomorphism. However, it's not hard to find two points $(a_1, b_1)$ and $(a_2, b_2)$ between the two lines that force $h$ to be decreasing somewhere.
If $\delta\geq 1$, then you will need $3$ points to ensure that $h$ is increasing some places and decreasing other places. Of course, if you allow, say, $b_1 = b_2$ with $a_1\neq a_2$, then that works regardless of $\delta$.