Yes your ideas are correct. For the sake of not leaving this question unanswered, I'll refine your proof a bit.
Proofs in graph theory have a tendency to naturally be broken into cases like your proof above. A lot of the time though, this isn't necessary, and you can make a proof more concise by trying to combine cases.
For example, in your proof, separating out the case where $n=2$ is not necessary. The inequalities in your $n > 2$ case covers it. In fact explicitly stating a base case is often only necessary when you are arguing by induction. Also separating the cases where there is one vertex of degree at most five and no vertices of degree at most five is not necessary. Since $6n \geq 6(n-1) + \operatorname{deg}(v_k)$, your work for the first cases actually covers both cases.
Theorem: For $n \geq 2$, a planar graph with $n$ vertices must have at least two vertices with degree at most five.
Proof $\;$ First note that if this theorem holds for connected graphs, then it must hold for non-connected graphs as well because we can consider a single component and find the required two vertices with degree at most five.
Let $G$ be a connected planar graph with $n \geq 2$ vertices and $e$ edges. Suppose for the sake of contradiction that $G$ has at most one vertex, call it $v_k$, of degree less than or equal to five. Then by the handshaking lemma we have
$$
2e = \!\!\!\sum_{v_i \in \operatorname{V}(G)}\!\!\!\operatorname{deg}(v_i)
\geq 6(n-1) + \operatorname{deg}(v_k)
\geq 6n-1\;.
$$
But since $G$ is planar and connected we know that $e \leq 3n-6$, getting us the contradiction
$$
6n-1 \;\leq\; 2e \;\leq\; 2(3n-6) = 6n-12\;.
$$
Therefore there must be at least two vertices of degree at most five.
The discharge method works here. (You can forget this proof, but try to remember the method.
It can be very powerful.
It belongs in your toolbox next to the pigeonhole principle and the principle of induction).
Assume $G$ is a counterexample with a minimum number of vertices ($n$).
We may also assume that we have maximized the number of edges ($e$).
So $G$ has minimum degree at least three and $d(u)+d(v)>13$ for every edge $uv$.
The edge maximization implies that if there are any non-adjacent $u$ and $v$ with $d(u)+d(v)>13$,
then $G+uv$ is not planar.
Now we assign a charge to each vertex: a vertex $v$ gets charge $6-d(v)$.
Applying the formula $e\leq 3n-6$ (equality only holds for triangulations),
we find the total charge $\sum_v(6-d(v))=6n-2e\geq6n-(6n-12)=12$.
Now we discharge: a vertex $v$ with a positive charge distributes its charge evenly among its neighbours.
We do this for all vertices simultaneously.
Since no charge is lost, there must still be vertices with a positive charge.
Let $v$ be a vertex with a positive charge.
Case 1: $d(v)\leq 8$. Since a vertex can only obtain a positive charge from a neighboring vertex $u$ with degree
at most 5, we find an edge $uv$ with $d(u)+d(v)\leq 13$. Contradiction.
Case 2: $d(v)=9$. $v$ has no neighbours of degree 3 or 4, so its positive charge can only be obtained
from neighbours of degree 5 and each of them contributes a charge of $\frac 15$
(they started with charge 1 and sent an equal part, i.e. $\frac 15$ to each of their neighbours).
Since $v$ started out with charge $-3$ it would need 16 neighbours to end up with a positive charge.
Contradiction.
Case 3: $d(v)=10$. As case 2: no neighbours of degree 3, and each neighbour of degree 4 or 5
contributes at most charge $\frac 12$. Since $v$ started with charge $-4$, it needs at least 9 neighbours
of degree 4 or 5. Let $v_1,\ldots,v_{10}$ be the neighbours of $v$, arranged clockwise.
Without loss of generality we may assume that $v_{10}$ is the only vertex that (possibly) has a degree larger than 5.
Since $v_1$ and $v_2$ cannot be adjacent (since $d(v_1)+d(v_2)\leq10$),
the face containing $v,v_1,v_2$ contains at least one other neighbour $u$ of $v_1$.
The degree of $u$ must be at least 9 (or $d(v_1)+d(u)\leq13$).
If $u$ is not adjacent to $v$, we can add edge $uv$, contradicting the edge-maximality. So $u$ is another neighbor of $v$, which implies $u=v_{10}$.
We have shown that $v_1$ is adjacent to $v_{10}$. Exactly the same argument shows that $v_2,\ldots,v_9$ are adjacent to $v_{10}$.
Now consider the adjacent "4-gons" $vv_2uv_1$ and $vv_2uv_3$. Since $d(v_2)\geq4$, $v_2$ must have neighbours that are inside one of these 4-gons, and not neighbours of $v$. Without loss of generality we may assume that $v_2$ has clockwise neighbours $w_0=v,w_1,\ldots,w_t,w_{t+1}=v_2$ where $t\geq1$ and $w_1,\ldots,w_t$ are inside $vv_2uv_1$. Since $d(w_1)\geq3$ we can now add edge $vw_1$ and again contradict edge-maximality.
Case 4: $d(v)\geq11$. Each neighbour of degree <6 contributes at most 1.
$v$ starts with charge $6-d(v)$, so it needs at least $d(v)-5$ neighbours of degree less than 6
to end up with a positive charge. Since $d(v)-5>\frac{d(v)}2$ we again must find
two consecutive neighbours of low degree and, like in case 3, these must either be
adjacent or expose a high-degree neighbour that can be connected to $v$ without
breaking planarity.
This final contradiction finishes the proof.
For the counterexample we can repeat the same arguments and find that there must be at least 12 vertices of degree 10, each of them surrounded by 5 vertices of degree 3.
This immediately brings up the idea of the icosahedron and indeed, if you take the icosahedron, and add a vertex in the center of each triangle face and connect it to the three corners, you get a very symmetric planar graph with 20 vertices of degree 3 and 12 of degree 10. No two vertices of degree 3 are adjacent, so this is our desired counterexample.
Edit to add: a stellated icosahedron:
Best Answer
Let $G$ be a planar graph with $n$ vertices and the minimum number $k$ vertices of degree at most $7$. We may assume that $G$ is maximal, since adding edges doesn't increase the number of "small degree" vertices.
In this case, if $n>3$ there are no vertices of degree two, since a path going through a degree two vertex can't be in two faces bounded by three edges.
From Euler's formula: $$6n - 12 \ge 3k + 8(n - k)$$ or $$5k \ge 2n + 12$$ If $n=12$ we discover $$5k\ge 36$$ and so $$k > 6$$