Algorithm to convert undirected connected graph with no bridges to strongly connected directed graph
Well the idea of dfs is quite good all you need to orient all edges away from the vertex that discovered them.

You will have a path from the root to each  vertex since the graph is connected and the dfs will visit all vertices.

Claim. From each vertex you can reach the root.

Proof. Let $\mathcal{C}$ be the strongly connected components of the resulting graph. Let $\mathcal{G}$ be the DAG resulting from contracting the vertices in each component in $\mathcal{C}$. Also, let $C_0 \in \mathcal{C}$ be the strongly connected component containing $r$. We have to prove that $C_0$ is the only component in $\mathcal{C}$ and hence, it contains all the vertices in the graph. Assuming otherwise, all components must be reachable from $C_0$ in $\mathcal{G}$, since all vertices are reachable from $r$ in $G$. choose $C$ to be a component in $\mathcal{C}$ that is the end of a longest path $P$ from $C_0$ in $\mathcal{G}$. Let $C'$ be the component previous to $C$ on that path. There is an edge from a vertex in $C'$ to a vertex in $C$. But this edge is not a bridge, hence, there must be an edge leaving $C$ and directed from $C$ to outside $C$ (since such an edge must have been discovered the first time we visited $C$). If this edge goes to a component $C''$ not on $P$, we could have extended $P$ to $C''$ resulting in a longer path wich is a contradiction. On the other hand, if it goes to a component on $P$, it forms a cycle, which contradicts the assumption that $\mathcal{G}$ is the DAG over the SCCs of the graph we build from $G$ and hence cannot have cycles. Hence, $\mathcal{C}$ cannot have more than one component, namely $C_0$.