We will count the outer face as its own face. So a cycle has two faces, for example.
The idea is to take each connected component $C_1$, $C_2$, etc., and connect them by drawing a bridge between $C_1$ and $C_2$, $C_2$ and $C_3$, etc. We will thus add $n-1$ edges (the restriction is if you contract the connected components we had previously, you must form a tree).
- $F = f$ remains the same.
- $V = n$ since no vertices were added.
- $E = m + k - 1$ since $k-1$ edges were added.
Now apply Euler's formula.
I found counting edge/faces this way very confusing. For example there are arguments such as: "there are at least 3 edges bounding a face and at most two faces sharing an edge then $3 | F| \le 2 | E|$". I assume that for $\sum_t F_t$ you mean |F|, the total number of faces in the graph $\Gamma(V,F)$. I do not see the $3 | F | \le 2 | E |$ obvious from the the "least" and "most" edge/faces statements above. Instead I prefer to put the problem in terms of triangles which are much simpler to understand. Any face can be decomposed in triangles. Another way (and this is direct and simpler) is shown at the end of this proof. This is based on adding all degrees (number of edges) for each face since they will count redundancies as well; then use that formula to link the number of edges with the number of faces in an inequality relationship. We prove:
Proposition: In a planar graph with two or more vertices,
$3 |F| \le 2 |E| $.
Proof:
We first assume that the faces are all triangles .
For example for one face (besides the exterior face) $|E|=3$ and $F=2$ so the equality $6=6$ is achieved. However for two faces
(for example a rectangle with a diagonal) we have $|E|=5$ and
$|F|=3$, here the inequality $9<10$ is strict.
We do this by induction over $|F|$ . We introduce a new face $|F|$ by
adding one more vertex and two more edges (note that since all faces are triangles we can not add a new face
in the interior of the graph). We need to show
that $2|E+2| \ge 3|F+1|$.
\begin{equation}
2 | E + 2| = 2 | E| + 4 \ge 3|F| + 4 \ge 3|F| + 3 \ge 3|F+1|.
\end{equation}
Let us now assume that at least one face $f$ is not triangular. Then we can divide the face into two new faces $f_1$ and $f_2$ with the help of some "diagonal" edge and form a graph $\Gamma'(F',V')$ which is the original $\Gamma(F,V)$ with the new split face. We can then consider a partition of set $F'$ into
$F'= F_1 \cup F_2$, such that $f_1 \in F_1$ and $f_2 \in F_2$. We apply an induction argument over the number $|F'|$. If $|F'|=2$, since a face has at least 3 edges then $2 | E | \ge 3 | F|$. Let us assume that the proposition is valid for any number $2 \le n \le | F|$. Since each graph $\Gamma_1(F_1, V_1)$, and $\Gamma_2(F_2, V_2)$ has less or equal number of faces than the graph $\Gamma'(F',V')$, we can apply the induction hypothesis to both $\Gamma_1$ and $\Gamma_2$. That is:
\begin{equation}
2 |E_1 | \ge 3 | F_1 | \quad \quad \mathrm{and} \quad \quad 2 | E_2 | \ge 3 | F_2 |.
\end{equation}
Now, $|E| = |E_1| + |E_2| -2 $ since the common edge was added twice,
and $|F| = |F_1 | + | F_2| - 2 $ since the outside (infinite face) was counted twice and one more face was counted in the interior. Then
\begin{eqnarray}
2 | E | &=& 2|E_1 | +2 |E_2| - 4 \\
&\ge& 3 |F_1| + 3 |F_2| -4 \quad \text{ by induction hypothesis} \\
&=& 3 | F_1 + F_2 - 2 | + 6 - 4 \\
&>&3 |F|.
\end{eqnarray}
In fact there is a more direct and simpler way to prove this:
Let us call $d(f_i)$ the degree of a face $f_i$. That is, the number of edges
bounding that face. we observe that
\begin{equation}
\sum_i d(f_i) \le 2 |E| \quad \quad \quad (1)
\end{equation}
since each edge that is part of a face contributes to exactly two faces, then add all those edges that are not part of a face.
On the other hand each face is bounded by at least 3 edges,
so $\sum d(f_i) \ge 3|F|$, and using equation (1) we find that
$2 |E| \ge 3 |F|$.
In general, and using exactly the same method and assuming that the
minimum number of edges of a face in a graph is $g \; (g\ge3)$ then we can write
\begin{equation}
2 | E| \ge g | F|.
\end{equation}
Best Answer
If you can 6-color each connected component, then you can 6-color the whole graph, by taking the union of the 6-colorings. So you only need to prove the theorem for a connected graph, and then it extends to unconnected graphs as a trivial corollary.