As you initially asked for how to approach this proof:
$(1)$ You can start by assuming $G$ is a connected planar graph having girth at least $6$.
Given such a $G$, then, you need to prove $G$ has at least one vertex with degree at most $2$.
$(2)$ You can also prove the statement in its equivalent contrapositive form:
Assume $G$ does not have a vertex with degree at most 2, and prove that then $G$ cannot be a connected planar graph having girth at least $6$. Proving this is equivalent to proving $(1)$.
(That is, you can prove any statement $p \rightarrow q$ by proving the contrapositive: $\lnot q \rightarrow \lnot p$, as these expressions are logically equivalent.)
$(3)$ Finally, you can use an indirect proof:
Assume that $G$ is a connected planar graph having girth at least $6$, AND assume, for the sake of contradiction, that $G$ does not have any vertex with degree at most $2$ (in other words, that for every vertex $v$ in $G$, the degree of $v\geq 3$).
With this approach, the task would then be to derive a contradiction. Upon reaching a contradiction, you can then conclude that if $G$ is a connected planar graph with girth at least $6$, then it cannot be the case that $G$ has no vertex of degree at most $2$. In short, it then follows that $G$ must have at least one vertex of degree at most $2$.
Whatever approach you use in your proof, you'll need to use the definition of a connected planar graph, and one that has girth at least $6$ (where the girth of a graph is the length of its shortest cycle).
You may also need to use Euler's formula as it relates to planar graphs: $$v-e + f = 2,$$ where $v$ is the number of vertices of $G$, $e$ the number of edges of $G$, and $f$ the number of faces. (In planar graphs, the "faces" are the number of regions bounded by edges, including the outer, infinitely large region.)
For the first part we can use Euler's $F-E+V=2$, where $F$ is the number of faces (counting the face that includes $\infty$), $E$ is the number of edges, and $V$ the number of vertices.
If there are no triangles then each face is enclosed by at least $4$ edges. So $4F/2<E$, i.e. $F<E/2$. From this we get $V>E/2+2$. If all vertices have order $\geq4$ then $4V/2<E$, i.e. $V<E/2$. From this we get $0>2$. So, we cannot have no triangles and at the same time all vertices with degree $\geq4$.
For the second part take a vertex that has degree $\leq3$. Color it and its neighbors with different colors $A,B,C,D$. We can do this only because it has degree $\leq3$. We can delete this vertex and the edges insident on it. If we color the rest of the graph there is always a way to color the deleted vertex. Notice that the graph with this vertex deleted still has no triangles, but less vertices. Apply induction on the number of vertices now.
Best Answer
The sum over all of the faces of the number of edges of each face is equal to twice the number of edges. Since the graph is triangle free this sum is at least $4F$ (Because every face has at least four edges) where $F$ is the number of faces. Hence $4F\leq 2E\implies F\leq \frac{E}{2}$ now plug this into Euler's formula:
$V-E+F=2\implies V-E+\frac{E}{2}\geq 2 \implies V-2\geq \frac{E}{2}\implies E\leq 2V-2$. Since the sum of the vertices twice the number of edges we obtain it is less than or equal to $4V-2$, so it is strictly less than $4V$, hence there must be a vertex of degree strictly less than $4$ as desired.