Is the alternative proof correct? if not why?

Question

(If $\ker f\subset \ker g$ where $f,g $ are non-zero linear functionals then show that $f=cg$ for some $c\in F$.) I came across such a task in the textbook only with the condition if $kerf= kerg$

$V = kerf + V/kerf$

Let's say there are two independent vectors in V/kerf
$e_1$ $e_2$

$f(e_1) \neq 0$ $f(e_2) \neq 0$

since $f$ functional there is a non-zero linear combination $x_1*f(e_1) + x_2*f(e_2)$ = $0 $ mean $f(x_1*e_1 + x_2*e_2) = 0$

$x_1 = 1 $

$x_2 = -f(e_1)/f(e_2)$

$x_1*e_1 +x_2*e_2 \neq 0$

$x = x_1*e_1 + x_2*e_2 $

x belongs kerf and V/kerf , contradiction

$dimV/kerf \leq 1$

and $f(x) = a*g(x) <=> f(e) = a*g(e)$ for any basis vector $e$ (Hamel basis) but we have maximum one basis vector for which f(x) is not zero

for a finite-dimensional space it can be shown that the dimension $dimkerf = n-1$ =>$dimV/kerf$ = $n-dimkerf$ = 1