If $\ker f\subset \ker g$ where $f,g $ are non-zero linear functionals then show that $f=cg$ for some $c\in F$.

Let $V$ be a vector space with $\dim V=n$ .

If $\ker f\subset \ker g$ where $f,g $ are non-zero linear functionals then show that $f=cg$ for some $c\in F$.

Now let $\mathcal B=\{v_1,v_2,\ldots ,v_n\}$ be a basis of $V$,

since $f,g$ are non-zero linear functionals then $\exists v_i\in \mathcal B $ such that $g(v_i)\neq 0\implies f(v_i)\neq 0$

Take $i=1$ without any loss of generality so take $g(v_1)\neq 0,f(v_1)\neq 0$.

Now take $c=\dfrac{f(v_1)}{g(v_1)}$

Then we need to show that $(f-cg)(v_i)=0\forall i$

Now $(f-cg)(v_1)=0$

How to show that $(f-cg)(v_i)=0\forall i\ge 2$

Can someone please help?

Note::Another Question Why do we need the dimension of the vector space to be finite?

There is no need for finite dimensionality and no need to use bases. Let $f(x) \neq 0$, $y$ be arbitrary and consider $y-\frac {f(y)} {f(x)} x$. By linearity we get $f(y-\frac {f(y)} {f(x)} x)=0$. By hypothesis this implies $g(y-\frac {f(y)} {f(x)} x)=0$. Hence $g(y)=cf(y)$ where $c=\frac {g(x)} {f(x)}$. Hypothesis implies that $c \neq 0$ so we can write $f=\frac 1 c g$.

Let me provide another approach: (it might be used to solve the problem with your last line but it is completely independent to your approach).

By definition $f,g:V\rightarrow\mathbb{R}$ are non-zero linear functionals (you can replace $\mathbb{R}$ with $\mathbb{C}$ or any field). By the rank-nullity theorem we have that $\dim \ker f = \dim \ker g = n-1$ Since $\ker f \subseteq \ker g$ we conclude that $\ker f = \ker g$. (In some sense this shows that $f-cg(v_i)=0$ in your solution because $f(v_i)=g(v_i)=0$.)

Now take a basis $v_1,...,v_{n-1}$ for the kernel and take $v$ which is linearily independent of those. Then $f(v),g(v)\not = 0$ are real numbers.

Take $c=\frac{f(v)}{g(v)}$. Since $f,g$ are non-zero only on $\text{span} ({v})$ the rest of the claim is immediate