By strong induction on $n = |V(G)|$. The base case $n \leq 5$ is trivial.

Step 1: Finding a low-degree vertex. By Euler's formula for planar graphs: $|E| \leq 3|V| - 6$. The average degree is $2|E|/|V| \leq 6 - 12/|V| < 6$, so there exists a vertex $v$ with $\deg(v) \leq 5$.

Step 2: Induction for $\deg(v) \leq 4$. Remove $v$; by induction, $G - v$ is 5-colorable. Since $v$ has at most 4 neighbors, at most 4 colors are used on them, leaving a color available for $v$.

Step 3: The case $\deg(v) = 5$. Let the neighbors of $v$ be $v_1, v_2, v_3, v_4, v_5$ in clockwise order around $v$ (using the planar embedding). By induction, $G - v$ has a proper 5-coloring. If the five neighbors use at most 4 colors, we can color $v$ with the fifth color. So assume all 5 colors $\{1,2,3,4,5\}$ appear on $v_1, \ldots, v_5$, with $c(v_i) = i$.

Step 4: Kempe chain argument. Consider the $(1,3)$-Kempe chain from $v_1$: the maximal connected subgraph of vertices colored 1 or 3 containing $v_1$.

Case A: $v_3$ is not in this Kempe chain. Then swap colors 1 and 3 in the chain containing $v_1$. This gives a valid coloring of $G - v$ where $v_1$ now has color 3, so color 1 is free among $v$'s neighbors. Color $v$ with 1.

Case B: $v_3$ is in the $(1,3)$-Kempe chain from $v_1$. Then there is a path from $v_1$ to $v_3$ using only colors 1 and 3. This path, together with $v$, forms a "barrier" separating $v_2$ from $v_4$ in the plane (by the Jordan curve theorem applied to the planar embedding).

Now consider the $(2,4)$-Kempe chain from $v_2$. Since $v_2$ and $v_4$ are separated by the $(1,3)$-path, $v_4$ cannot be in the $(2,4)$-chain from $v_2$. Swap colors 2 and 4 in the chain containing $v_2$. Now $v_2$ has color 4, freeing color 2. Color $v$ with 2.

In all cases, $G$ is 5-colored. $\square$