Let $G$ be strongly connected directed graph with vertex set $\{1, 2, \ldots, n\}$. A circuit $i_0=i, i_1, \ldots, i_n=i$ is elementary if path $ i_1, \ldots, i_{n-1}$ is a path in which no vertex appears more than once.
In a paper, author claims that if the length of any elementary circuit of $G$ is larger than or equal to 2, then there is a set $S=\{i_1,_2,\ldots, i_m\}\subseteq \{1, 2, \ldots n\}$ with $m\leq [\frac{n}{2}]+1$ such that any elementary circuit of $G$ contains at least one vertex that belongs to $S$.
I do not know proof of it. Can you please help me to know it?
Indeed proof of the following lemme is not clear for me.
Lemma. Let $G$ be strongly connected directed graph with vertex set $\{1, 2, \ldots, n\}$ and the length of any elementary circuit of $G$ is larger than or equal to 2. If $i,j\in\{1, \ldots n\}$, then there exists a (possibly empty) path $P_{ij}$ from $j$ to $i$ in $G$ that passes through at least one vertex of each (elementary) circuit of $G$ and that has a length that is less than or equal to $\frac{n^2-1}{2}$.