Perfect matching theorem for a cubic graph with at most two bridges

Every finite connected cubic graph with at most two bridges has a perfect matching.

Proof. The handshaking lemma implies that the number $n$ of vertices is an even integer. Fix a nonempty subset $S$ of the vertex set, and let $q$ count the components of $G-S$ of odd integer order. For such a component $C$, counting vertex degrees gives $3|C|=2|E(C)|+d(C)$, where $d(C)$ counts its edges to $S$. Thus $d(C)$ is a positive odd integer. If it is one, the unique outgoing edge is a bridge; this happens for at most two components. For all others $d(C)\ge 3$. Therefore $3q-4\le 3|S|$. Meanwhile $q\equiv n-|S|\equiv |S|(\mathrm{mod}2)$. If $q>|S|$, the same parity would force $q\ge |S|+2$, contradicting the preceding bound. For $S=\varnothing$ there is one component of even integer order and hence $q=0$. Apply Tutte's theorem on perfect matchings.