Proving that every third Fibonacci number is divisible by F2=2

There's a missing term in (B) as you noticed. A correct proof is by induction. Show that $F_2$ is even (immediate). Then write

$$ \begin{align} F_{3n+2} &= F_{3n+1} + F_{3n} \\ &= 2F_{3n} + F_{3n-1} \\ &= 2F_{3n} + F_{3(n-1)+2} \enspace, \end{align} $$

which is even because it's the sum of two even numbers. ($F_{3(n-1)+2}$ is even by the induction hypothesis.)

There is a well-known property of Fibonacci numbers: $$ \gcd(F_n,F_m)=F_{\gcd(n,m)} \tag{1} $$ Since $F_3=2$, $$ \gcd(F_{3n},2)=\gcd(F_{3n},F_3)=F_{\gcd(3,3n)} = F_3 = 2 \tag{2} $$ hence every Fibonacci number of the form $F_{3n}$ is even.

Base case:

$F_1=1\\ F_2=1\\ F_3=2$

Inductive hypothesis: Suppose $F_{3k-2}$ is odd, and $F_{3k-1}$ is odd, and $F_{3k}$ even.

We must show that $F_{3(k+1)-2}$ and $F_{3(k+1)-1}$ are odd, and $F_{3(k+1)}$ is even

$F_{3(k+1)-2} = F_{3k-1}+F_{3k}$

Based on the inductive hypothesis, $F_{3(k+1)-2}$ must be odd as the sum of an odd number and an even number are odd.

I will leave it to you to finish from here.