Let $n$ be the number of vertices of $G$, $m$ the number of edges of $G$, and $r$ the number of faces of $G$. We use the notation $\mathcal{E}(f)$ for the number of edges on the face $f$.
We will assume that $G^*$ has no vertex of degree $3$ or less, and prove that $G$ must then have a vertex of degree at most $3$.
If $G^*$ has no vertex of degree $3$ or less, than every face of the original graph $G$ has at least $4$ edges on its boundary.
Since every edge appears twice on a face boundary (either once on two different faces, or twice on one face if it is a cut edge), we get that:
$$\sum_{f\text{ a face of $G$}} \mathcal{E}(f) = 2m$$
We have assumed that $G^*$ has minimum degree at least $4$, so this gives:
$$2m = \sum_{f\text{ a face of $G$}} \mathcal{E}(f) \geq 4r$$
By substituting $r\leq \frac{m}{2}$ into Euler's Formula ($n-m+r = 2$), we get that:
$$2 = n-m+r \leq n-m + \frac{m}{2}$$
Rearrange this to bound the number of edges as $m\leq 2n-4$.
Since $2m = \sum_{v\in V(G)}d(v)$, we get that the sum of the degrees is bounded by:
$$\sum_{v\in V(G)}d(v) = 2m \leq 4n-8 < 4n$$
The sum of the degrees of $G$ is less than $4n$, so the average degree of $G$ is less than $4$ (the average degree is the sum of the degrees divided by $n$). This means there is at least one vertex with degree less than $4$, completing the proof.
Take a straight line embedding of the maximal planar graph. By maximality every face is a triangle. To show that it is $2$-connected, take any two faces $F_1$ and $F_2$, and any two points $p$ and $q$ in the interior of $F_1$ and $F_2$ respectively, and consider the line $(pq)$. For almost any choice of $p$ and $q$, this line does not intersect any vertex, and intersect each face on a (possibly empty) segment, except the outer face for which the intersection is the complement of a segment. Hence the intersected faces defines a circuit in the dual graph, containing $F_1$ and $F_2$ (as well as the outer face). As the connectivity between any two faces is at least 2, the graph is 2-connected, hence 2-edge-connected.
This also shows that your intuition is right: we can find two paths from every face to the outer face in the same way.
Best Answer
The dual graph need not be simple. Here’s a badly drawn example of a black tree with $6$ vertices (and therefore $5$ edges and $1$ face) and its red dual: the dual has $1$ vertex, $5$ edges, all of which are loops, and $6$ faces.