For fixed $u,v \in V$ consider the set
$L(u,v) = \{ f \in \mathbb{R}^V: \pi_u(f) + \pi_v(f) = \pi_{u+v}(f)\}$. Letting $p: \mathbb{R}^V \to \mathbb{R}^3$ be the map sending $f$ to $(f(u), f(v), f(u+v))$, which is continuous as a "triple projection", note that $L(u,v) = p^{-1}[H]$ where $H$ is the hyperplane $\{((x,y,z) \in \mathbb{R}^3: x+ y -z= 0\}$, which is closed in $\mathbb{R}^3$, so $L(u,v)$ is always closed.
Similarly, for each scalar $\alpha \in \mathbb{R}$ and $u \in V$ the set
$M(\alpha, u) = \{f \in \mathbb{R}^V: \pi_{\alpha u}(f) = \alpha \pi_u(f)\}$ can be seen as $p'^{-1}[L]$ where $p': \mathbb{R}^V \to \mathbb{R}^2$ is given by $p'(f) = (f(\alpha u), f(u))$ and $L = \{(x,y) \in \mathbb{R}^2: x = \alpha y \}$ is closed. So all $M(\alpha, u)$ are closed too.
Now, denoting by $\pi_B$ the restriction map from $\mathbb{R}^V$ to $\mathbb{R}^B$, $$B^\ast = \pi_B\left[ [0,1]^V \cap \bigcap\{ L(u,v): u,v \in B, u+v \in B\} \cap \bigcap \{M(\alpha,u): u \in B, \alpha u \in B\} \right]$$ which is closed in $[0,1]^B$, as the set inside the $[]$ is compact (closed in $[0,1]^V$); intersections of closed sets are closed after all.
The crux is seeing "being a functional" as having to obey a lot of "closed conditions".
Some choice is needed, since it is consistent with $\mathsf{ZF}$ to have a nontrivial vector space $V$ such that $V^\ast=\{0\}$, and I'm talking about the algebraic dual here.
With choice given $e\in V$ with $e\neq 0$ it's easy to come up with a functional which is not zero on $e$: extend $\{e\}$ to a basis $\{e\}\cup\{v_i\mid i<\kappa\}$ of $V$ and define $\phi:V\to\Bbb R$ by $\phi(e)=1$ and $\phi(v_i)=0$ for every $i$. Extend by linearity to the whole space.
Note that in the context of Banach spaces we can even come up with a continuous functional which is nonzero on $e$, by using Hahn-Banach to extend the functional $\phi\colon\langle e\rangle\to\Bbb R$ given by $\phi(ae)=a$ to a bounded functional on the whole space.
Best Answer
Define $T: X\to \mathbb R^{n}$ by $T(x)=(f_1(x),f_2(x),...,f_n(x))$. If $b =Tx$ for some $x$ then, for any linear map $g:\mathbb R^{n} \to \mathbb R $, $g(Tx)=0$ implies $g(b)=0$. This proves one way. [Just note $g(t_1,t_2,...,t_n)$ is of the form $\sum \lambda_i t_i$].
For the other direction we have to use the fact that if $b$ is not in the range of $T$ then there exists a linear map $g:\mathbb R^{n} \to \mathbb R $ such that $g=0$ on the range of $T$ but $g(b) \neq 0$.