Try proving the contrapositive of the $\impliedby$ direction instead:
If $G$ is not a forest, then there is some induced subgraph of $G$ whose vertices all have degree at least $2$.
Indeed, since $G$ is not a forest, it contains a cycle. Now let $H$ be the subgraph of $G$ induced by the vertices of this cycle. Then since $H$ contains a cycle, each of its vertices has degree at least $2$, as desired.
Suppose $S$ has a separating set $S = \{v,w\}$. Let $C$ be the smallest component of $G-S$ that doesn't have any vertices of degree $\le 2$, and assume that $S$ is chosen to make $C$ as small as possible.
This, in particular, means that each of $v$ and $w$ has at least two neighbors in $C$. If, say, $w$ has only one neighbor $w'$ in $C$, then $S' = \{v,w'\}$ is another separating set of size $2$, and one of the components of $G-S'$ is $C-\{w'\}$, which is smaller than $C$. (And $C - \{w'\}$ cannot be empty, because then $w'$ would have had degree $\le 2$.)
Also, assuming $G$ is $2$-connected, there is a path connecting $v$ and $w$ outside $C$. Let $x$ be a neighbor of $v$ outside $C$; deleting $v$ should leave a path from $x$ to $C$, which must go through $w$.
Now create a new graph $G'$ in which all vertices outside $C \cup S$ are deleted, and we add a new edge $vw$ if it did not exist already. In $G'$, $v$ and $w$ still have degree at least $3$, and the degrees in $C$ are unchanged.
By induction, $G'$ contains a subdivision of $K_4$. This can be turned into a subdivision of $K_4$ in $G$: if the edge $vw$ is needed and not present in $G$, we can replace it by a path from $v$ to $w$ outside $C$, which we know exists.
Alternatively, if you're willing to use some powerful classification results similar to your observation about planar graphs, then you can start from the fact that the graphs with no subdivision of $K_4$ are precisely the series-parallel graphs.
A series-parallel graph must have a vertex of degree $2$. More precisely, a series-parallel graph with source $s$ and sink $t$ either has no other vertices, or else has a vertex of degree $2$ other than $s$ and $t$. To prove this, you can verify that this property is preserved under both series composition and parallel composition:
- A series composition of two $K_2$s results in a path on $3$ vertices, which creates a vertex of degree $2$ other than $s$ and $t$. A parallel composition of two $K_2$s creates a multigraph, so it's not allowed.
- In a series or parallel composition of two graphs, at least one of which has three or more vertices, one of the graphs already had a vertex of degree $2$ other than $s$ and $t$. That vertex still has degree $2$ after the composition.
Therefore a graph with minimum degree $3$ cannot be series-parallel, so it has a subdivision of $K_4$.
We can refine this argument to also rule out graphs in which only one vertex has degree less than $2$. If our series and parallel composition involved at least two graphs with three or more vertices at any point, then each of them had an internal vertex of degree $2$. If not, then $s$ and $t$ can each have degree at most $2$: their degrees increase only with parallel compositions, and we can't parallel-compose a $2$-vertex graph with itself, so for $s$ or $t$ to reach degree $3$, we'd need two larger graphs.
Best Answer
"A planar triangulation is 4-connected if and only if it has no separating triangle."
$\Rightarrow$:
If there is a separating triangle then there is a 3-cut set. The graph is therefore not 4-connected.
$\Leftarrow$:
A planar triangulation is 3-connected. If the graph is not 4-connected, then any minimal cutset $S$ is a set of 3 vertices. A planar triangulation is a chordal graph. And in a chordal graph, any minimal cutset is a clique. So $S$ is a separating triangle.