A Berge cycle bounds the union of its hyperedges

Put $U={\bigcup }_{i<t}{e}_i$ and $I=\{0,\dots ,t-1\}$. These are finite, with $\mathrm{\mid }I\mathrm{\mid }=t$. Define $R=\{(v,i)\in U\times I:v\in e_i\}$ and $m(v)=\mathrm{\mid }\{i\in I:v\in e_i\}\mathrm{\mid }$ for $v\in U$. The row fibre at $v$ has size $m(v)$ and the column fibre at $i$ is $e_i$, of size $r$. Double counting therefore gives ${\sum }_{v\in U}m(v)=\mathrm{\mid }R\mathrm{\mid }={\sum }_{i\in I}r=rt$.

Every $v\in U$ lies in at least one displayed edge, so $m(v)\ge 1$ and $d(v)=m(v)-1$ is a natural number satisfying $m(v)=1+d(v)$. Additivity and the constant-summand law yield $rt={\sum }_{v\in U}(1+d(v))=\mathrm{\mid }U\mathrm{\mid }+{\sum }_{v\in U}d(v)$.

Each cycle vertex $x_i$ belongs to the two distinct edges $e_i,e_{i+1}$, so $d(x_i)\ge 1$, also when $t=2$. Enumerate $U$ with the $t$ distinct cycle vertices first. By splitting the sum after these vertices, monotonicity and the constant-summand law give ${\sum }_{v\in U}d(v)\ge t$, since the remaining terms are nonnegative. Thus $\mathrm{\mid }U\mathrm{\mid }+t\le rt=(r-1)t+t$, and cancellation gives $\mathrm{\mid }U\mathrm{\mid }\le (r-1)t$.