What is the minimal pumping length of this string $(01)^*$
The pumping length of a regular language $L$ is the minimal $p$ such that every word $w \in L$ of length at least $p$ can be split as $w = xyz$ such that (i) $|xy| \leq p$, (ii) $y \neq \epsilon$, (iii) $xy^iz \in L$ for every $i \geq 0$. The pumping length of $(01)^*$ using this definition is $2$. Anybody claiming that the pumping length is $1$ is either wrong or has a different definition in mind (the notion of pumping length isn't completely standard, and my definition is not the only possible one).

In order to show that the pumping length cannot be $1$, you need to show that there exists some word $w \in L$ of length at least $1$ such that for every splitting $w = xyz$ in which $|xy| \leq 1$ and $y \neq \epsilon$ there exists $i$ such that $xy^iz \notin L$. In particular, you don't get to pick the decomposition. You need to show that no decomposition works. In this case, we can take the word $w = 01 \in L$. Since all words in $L$ have even length, if $w = xyz$ and $xz \in L$ then necessarily $y$ has even length. However, this is impossible since on the one hand $|y| \leq |xy| \leq 1$, and on the other hand $y \neq \epsilon$.