If the bridges of a $3$-regular graph $G$ lie on a single path then $G$ has a $1$-Factor (perfect matching)
I've proved that a $3$-regular graph with at most two bridges has a perfect matching ($1$-factor), and I also know that a bridgeless $3$-regular graph has a $1$ factor, however I'm a bit lost here and I don't really see how to use the fact that the bridges lie on a single path. I tried to look at the $2$-connected components of $G$ and try to add a vertex to make each of these blocks $3$-regular so that each block has a $1$ factor but I think this doesn't really work. So any help would be greately appreciated.
Proceed by induction on the number of bridges. If there are at most two bridges, we are done. Suppose now that $G$ has $k\geq 3$ bridges, say $b_1, \dots, b_k$, labelled in order of the path. Deleting these yields subgraphs $C_1, \dots, C_{k+1}$, with the natural labelling.
Construct two new graphs, to which we will apply the induction hypothesis, as follows. Let $H$ be obtained from $K_4$ by subdividing an edge. Let $G'$ be obtained from $G[\cup_{i=1}^{k-1}V(C_i)]$ by adding a copy of $H$ and an edge between the vertex of degree two in the copy of $H$ and the vertex of degree two in $C_{k-1}$. Analogously, let $G''$ be obtained from $G[\cup_{i=k}^{k+1}V(C_i)]$ by adding a copy of $H$ and an edge between the vertex of degree two in the copy of $H$ and the vertex of degree two in $C_{k}$. Essentially $G'$ is $G$ where we have replaced the last two subgraphs by a copy of $H$, and $G''$ is $G$ where we replaced the first $k-2$ subgraphs by a copy of $H$.
Now, $G'$ has $k-1$ bridges on a path, $G''$ has $2$ bridges, and both $G'$ and $G''$ are cubic. By induction, $G'$ has a perfect matching $M'$ and $G''$ a perfect matching $M''$. Moreover, since $H$ has an odd number of vertices, $M'$ and $M''$ both contain the bridge to their respective copies of $H$. Thus, the matching $M$ obtained by taking all edges in $M'$ not in the copy of $H$, together with all edges in $M''$ not in the copy of $H$, and $b_{k-1}$, forms a perfect matching of $G$, as desired.
I believe this a theorem due to Alfred Errera from 1922, but I do not know if this is the proof he gave.