I am following the the book "Lecture notes on functional analysis". The following extension thm has been proven in the book.
If we consider the finite dimensional case of the $X$, does the following prove work for the existence of the linear functional $F$? $\to$ My guess is yes but I have some problem of proving the property $\|F\|=\|f\|$.
Let $\{x_1,\dots, x_k\}$ be basis for $V$. We can extend it to a basis $\{x_1,\dots, x_k,x_{k+1},\dots , x_n\}$ of $X$ and set $F(x_m)=0$ for $m = k+1, \dots , n$. So for any $X\ni x = a_1x_1+\dots+a_nx_n$ , we can define $F(x)=a_1f(x_1)+\dots+a_kf(x_k)$. Then $F$ is a linear functional on $X$, $F(x)=f(x) , \forall x\in V$, and $\|F\|=\|f\|$.