The following article provides a positive answer to my question. Yes, it is possible to prove that a 4-connected planar graph has a perfect matching or almost perfect matching even without using the Hamiltonian property.
- Biedl, Therese, et al. "Tight bounds on maximal and maximum matchings." Discrete Mathematics 285.1-3 (2004): 7-15.
Here is my selective excerpt (because the authors also did something else in the same article).
1. First, the authors introduce the concept of the 4-block tree.
Similar to the 2-block tree, we can define a 4-block tree that
captures the relationships among the 4-connected components of a graph
(Fig. 2). Recall that a graph is 4-connected if removing any three
arbitrary vertices leaves a connected graph. Assume that a graph is
3-connected, but not 4-connected. Then it contains three vertices
$\{v, w, x\}$ such that removing them from the graph yields at least
two connected components; we call $\{v, w, x\}$ a separating triplet.
For each connected component $C$ obtained from removing $\{v, w, x\}$,
we create a new graph by adding to $C$ the vertices $v, w, x$, as well
as all their edges incident to another vertex in $C$, and the three
edges $(v, w),(w, x)$ and $(x, v)$ if they did not exist already.
We iterate this process until all resulting graphs are 4-connected;
these are the 4-connected components of the graph. The 4-block tree is
then defined as follows. We create one node for every 4-connected
component, and one node for every separating triplet, and add an edge
if and only if the separating triplet was contained in the 4-connected
component. The resulting graph is again a tree. We denote its number
of leaves by $\ell_4(G)$, or just $\ell_4$ if the graph is clear from
the context.
Note that each leaf of the 4-block tree corresponds to some subgraph
of $G$ that would be 4-connected if we added all edges between the
vertices of the separating triplet that defined it.
2. Main theorem.
Nishizeki and Baybars showed that every 3-connected planar graph has a matching of size $\frac{n+4}{3}$ [9]. In this section, we
strengthen this result by including the number of leaves of the
4-block tree in the bound; in particular we obtain a bound that
resolves to $\left\lfloor\frac{n}{2}\right\rfloor$ if the graph is
4-connected.
Theorem 3. Any 3-connected planar graph $G$ of order $n$ has a matching of size $\min \left\{\frac{n-1}{2}, \frac{2 n+4-\ell_4}{4}\right\}$, where $\ell_4$ is the number of leaves of the 4-block tree of $G$.
Proof. Let $G$ be a 3-connected planar graph of order $n$, and let $M$ be a maximum matching in $G$. By Theorem 2 , there exists a vertex set $T$ in $G$ such that there are exactly $\left|V_{\neg M}\right|=\operatorname{odd}(T)-|T|$ unmatched vertices in $M$. If $|T| \leqslant 2$, then $G-T$ is still connected, i.e., $\left|V_{\neg M}\right| \leqslant \operatorname{odd}(T) \leqslant 1$. But then clearly $|M| \geqslant \frac{n-1}{2}$.
If $|T|=3$, then there can be at most two odd components in $G-T$. If there were three or more components, they would all have to be incident to all vertices of $T$ by 3 -connectivity, and the graph would contain $K_{3,3}$ as a minor. But $G$ is planar, so this is impossible. Since we assumed that there are $\operatorname{odd}(T)-|T|<0$ unmatched vertices, this case is actually impossible.
If $|T| \geqslant 4$, then we greedily add edges between any two non-adjacent vertices of $T$ that lie on the same face of $G$, without destroying the planarity of the graph. Let $G_T$ denote the subgraph of this augmented graph induced by the vertices of $T$ (see Fig. 3). Note that no two components of $G-T$ can be within the same face of $G_T$, because then we would have introduced an edge to split the face between them. Therefore, for every odd component there must be a unique face in $G_T$. This immediately proves $\operatorname{odd}(T) \leqslant 2|T|-4$, but in fact, we can do better and show $2 \operatorname{odd}(T) \leqslant 2|T|-4+\ell_4$.
More precisely, let $f_3$ and $f_{\geqslant 4}$ be the number of faces of $G_T$ of degree 3 and degree at least 4 , respectively. An easy counting argument shows that $f_3+2 f_{\geqslant 4} \leqslant 2|T|-4$. Let $C$ be an odd component, and let $f_C$ be the face of $G_T$ containing $C$. If $f_C$ has degree 3 , then $C$ has only three neighbors in $T$, and these three neighbors form a separating triplet of $G$ (separating $C$ from the rest of $T$, remember that $|T| \geqslant 4$ ). This separating triplet is the ancestor of at least one leaf of the 4-block tree of $G$. So $C$ can be associated with one face of $G_T$ that has degree 3 and one leaf of the 4-block tree. If $f_C$ has a higher degree, then $C$ can be associated with one face of $G_T$ that counts towards $f_{\geqslant 4}$. So $2 \operatorname{odd}(T) \leqslant f_3+\ell_4+2 f_{\geqslant 4} \leqslant 2|T|-4+\ell_4$.
But then $\left|V_{\neg M}\right| \leqslant \frac{2|T|-4+\ell_4}{2}-|T|=\frac{\ell_4-4}{2}$, which implies $\left|V_M\right| \geqslant n-\left|V_{\neg M}\right| \geqslant \frac{2 n+4-\ell_4}{2}$.
Regarding Question 3, here is a proof that $f(n)=30$ for all $n \geq 40$. Let $H$ be a $2$-connected planar graph with minimum degree $5$. Let $G$ be obtained from $H$ by adding a new vertex inside each face of $H$, and making it adjacent to all vertices of the face. Let $V(G)=X \cup Y$, where $X=V(H)$, and $Y$ are the newly added vertices. Since $H$ has minimum degree $5$, $\deg_G(x) \geq 10$ for all $x \in X$. Moreover, $\deg_G(y) \geq 3$ for all $y \in Y$ and no two vertices of $Y$ are adjacent. Thus, $\deg(u)\deg(w) \geq 30$ for all $uw \in E(G)$. Note that $G$ is a planar graph with $|V(H)|+|F(H)|$ vertices, where $F(H)$ is the number of faces of $H$. By Euler's formula, $|V(G)|=|E(H)|+2$. Since there is a $2$-connected planar graph with minimum degree $5$ and $m$ edges for all $m \geq 38$, we are done.
Best Answer
According to Frank Lutz's article it's in paragraph 46 of the Steinitz-Rademacher book: "every triangulated 2-sphere can be reduced to the boundary of the tetrahedron by a sequence of edge contractions".