For finite edges this follows indeed from compactness. Consider the Cartesian product $K=\prod e$ of all edges in Tychonoff topology. This $K$ may be understood as the set of simultaneous choices $v_e\in e$ for all edges $e$. For every pair $e\ne f$ of edges consider the open set $U(e,f)\subset K$ determined as follows: a vertex chosen in $e$ belongs to $f$. If these open sets do not cover $K$, then a choice set exists. If they do cover $K$, then finitely many if them cover $K$, that means that the choice set does not exist for finitely many edges.

For finite edges this follows indeed from compactness. Consider the Cartesian product $K=\prod e$ of all edges in Tychonoff topology. This $K$ may be understood as the set of simultaneous choices $v_e\in e$ for all edges $e$. For every pair $e\ne f$ of edges consider the open set $U(e,f)\subset K$ determined as follows: a vertex chosen in $e$ belongs to $f$. If these open sets do not cover $K$, then a choice set exists. If they do cover $K$, then finitely many of them cover $K$, that means that the choice set does not exist for finitely many edges.

For finite edges this follows indeed from compactness. Consider the Cartesian product $K=\prod e$ of all edges in Tychonoff topology. This $K$ may be understood as the set of simultaneous choices $v_e\in e$ for all edges $e$. For every pair $e\ne f$ of edges consider the open set $U(e,f)\subset K$ determined as follows: we choose the same vertex in $e$ and in $f$. If these open sets do not cover $K$, then a choice set exists. If they do cover $K$, then finitely many if them cover $K$, that means that the choice set does not exist for finitely many edges.

For finite edges this follows indeed from compactness. Consider the Cartesian product $K=\prod e$ of all edges in Tychonoff topology. This $K$ may be understood as the set of simultaneous choices $v_e\in e$ for all edges $e$. For every pair $e\ne f$ of edges consider the open set $U(e,f)\subset K$ determined as follows: a vertex chosen in $e$ belongs to $f$. If these open sets do not cover $K$, then a choice set exists. If they do cover $K$, then finitely many if them cover $K$, that means that the choice set does not exist for finitely many edges.