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.
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.
Best Answer
As @Licheng noticed, we need to be clear on the requirements for the graph $G$. From the similar questions mentioned by @Licheng, $G$ is supposed to be simple.
The proof is not that hard, but there are a lot of details to make it rigorous.
We start with some results, stated without proof:
Theorem : The cycle space of a planar graph $G$ is the cut space of its dual $G^*$
Corollary 1: $G$ has a self loop $\iff$ $G^*$ has a $1$-edge cut (a bridge).
Corollary 2: $G$ has an edge of multiplicity $\geq 2$ $\iff$ $G^*$ has a $2$-edge cut.
Corollary 3: $G$ is simple $\iff$ $G^*$ is $3$-edge connected.
We can then show:
Theorem 1: If $G$ is simple and $2$-vertex connected, then it implies $G^*$ is $2$-vertex connected and has no loops.
proof: We show it by contraposition:
If $G^*$ has a loop, then $G = G^{**}$ has a $1$-edge cut, hence it has a $1$-vertex cut, so $G$ is not $2$-vertex connected.
If $G^*$ has a $1$-vertex cut (and no loop), let $u$ be the cut vertex, and let $G_1$ and $G_2$ be two disconnected subgraphs that appears after removing $u$.
$u$ must be connected to at least one $G_1$ vertex, and one $G_2$ vertex. There is at least one face $f$ with the pattern $G_1-u-G_2$. If we follow the vertices of the face from $u$ in the $G_1$ direction, we can't go directly from a vertex of $G_1$ to a vertex of $G_2$, because these graphs are disconnected, so we will need to meet again $u$ before reaching $G_2$. So $f$ has the following pattern: $... G_2 - u - G_1 - ... - G_1 - u - ...$. We will denote $CG_1$ the connected component of $G_1$ containing the vertices of $G_1$ appearing in this pattern, and $CG_2$ the connected component of $G_2$ containing the vertex of $G_2$ appearing in this pattern.
If the graph induced by the vertices of $CG_1$ and $u$ has no faces (besides $f$), then it must be a tree and it will have a bridge, so $G$ will have a loop. We apply the same line of reasoning to $CG_2$
We can know assume the graph induced by the vertices of $CG_1$ and $u$ has at least a face, say $f_1$. We can also assume the graph induced by the vertices of $CG_2$ and $u$ has at least a face, say $f_2$. When deleting vertex $f$ in $G$ (faces are vertices in the dual), $f_1$ and $f_2$ will be disconnected, so $f$ is a cut-vertex of $G$, so $G$ is not $2$-connected.
Theorem 2: If $G$ is simple and $3$-vertex connected, then equivalently $G^*$ is simple and $3$-vertex connected
proof: We show it by contraposition:
If $G^*$ is not simple, then $G^*$ has a $1$ or $2$-edge cut, hence it has a $2$-vertex cut so $G$ is not $3$-vertex connected.
If $G^*$ has a $2$-vertex cut, let $u$, $v$ be these two cut vertices, and let $G_1$ and $G_2$ be two disconnected connected graph that appears after removing $u$ and $v$.
In the first time, we will suppose that there is no edge between $u$ and $v$, so $u$ and $v$ have only neighbors in $G_1$ or $G_2$.
If there is a face with a pattern like $...-G_1-u-G_2...-G_2-u-...$, we can apply the same reasoning than in theorem $1$ to prove that there is a one vertex cut, so we can now assume no such pattern exist.
Both $u$ and $v$ needs to be connected to both $G_1$ and $G_2$. By turning clockwise on the neighbors, we will find both a vertex of $G_1$ followed by a vertex of $G_2$, defining a face $f_1$, and a vertex of $G_2$ followed by a vertex of $G_1$ defining a face $f_2$. For a face with the pattern $G_1-u-G_2$, by following the contour of the face, we get the pattern: $u-G_1-...-G_1-v-(...-v)^*-G_2-...-G_2-u$.
By combining the patterns of $f_1$ and $f_2$, we then get the pattern:
We define $CG_1$ the graph induced by $u$, $v$ and the vertices of the connected component of $G_1$ containing the $G_1$'s vertices of $f_1$ and $CG_2$, and we define $CG_2$ similarly.
If $CG_1$ contains no faces, then $u$ and $v$ must have only one neighbors in $CG_1$, and this defines a $2$-edge cut. Same applies for $CG_2$.
If both $CG_1$ and $CG_2$ have a face, these faces are separated by $f_1$ and $f_2$, thus it defines a $2$-vertex cut in $G^*$, so a multi-edge in $G$
Now if there is an edge between $u$ and $v$, we delete it, and applies the above reasoning. Adding it back does not discard the reasoning on pattern $...-G_1-u-G_2...-G_2-u-...$, nor when $CG_1$ or $CG_2$ has no faces. Otherwise, $u-v$ split may split $f_1$ or $f_2$ in two parts, but we can keep only one of these two parts, and it will still define a $2$-vertex cut.