The closure is not the whole space. The fact that neighbourhoods contain lines is irrelevant, since the closure contains no neighbourhood.
Now let me show below that the intersection of the kernels of a finite set of functionals has to be non-trivial if $X$ is infinite-dimensional.
If you have functionals $f_1,\ldots,f_n$, consider the bounded linear map $\tilde f:X\to\mathbb C^n$ given by $\tilde f(x)=(f_1(x),\ldots,f_n(x))$. Then $\ker\tilde f=\bigcap_k\ker f_k$. So, if $\bigcap_k\ker f_k=\{0\}$, we would have $\tilde f$ linear and injective; it would then be an injection into a finite-dimensional space, which would imply $X$ is finite-dimensional. In conclusion, $\bigcap_k\ker f_k\ne\{0\}$
Take $V=K$ and $W=K^2$; consider $f\colon V\to W$ given by $f(1)=(1,0)$ and $f_1\colon V\to W$ given by $f_1(1)=(0,1)$. Of course $\ker f_1\subseteq\ker f$, but $f$ is not linearly dependent on $f_1$.
Let's look at the general case. If $U=\bigcap_{i=1}^n\ker f_i$, then your maps induce linear maps $\bar{f}_i\colon V/U\to W$ $(i=1,\dots,n)$ and $\bar{f}\colon V/U\to W$. The problem is now whether any linear map $g\colon V/U\to W$ can be obtained as a linear combination of the given ones.
Thus we can assume $U=0$ and that the maps are linearly independent, so the problem becomes
Let $f_1,\dots,f_n\colon V\to W$ be linearly independent maps of $K$-vector spaces with $\bigcap_{i=1}^n\ker f_i=0$; is it true that any linear map $f\colon V\to W$ is a linear combination of the given maps?
What we can say is that there is an embedding
$$
V=\frac{V}{\bigcap_{i=1}^n\ker f_i}\to\bigoplus_{i=1}^n \frac{V}{\ker f_i}.
$$
so $\dim V\le n\dim W$. Therefore, increasing the dimension of $V$ in the original problem doesn't help in the "normalized" situation. This shows also that the assertion is, in general, false: the span of the given maps has dimension $n$, while $\dim\mathrm{Hom}(V,W)=(\dim V)(\dim W)$. You should be able to show a counterexample, now.
Notice that, in the case when $W=K$, it suffices to show that the span of the given map has dimension $\dim V$, which follows from the above embedding.
Best Answer
Extend $e_1, \ldots, e_d$ to a basis $e_1, \ldots, e_n$ of $F_q^n$. For each $i$ between $1$ and $n$, define a linear form $e_i^*$ by its action on the basis: let $e_i^*(e_j)$ be $0$ when $i \neq j$ and $1$ when $i = j$. The linear forms $e_1^*, \ldots, e_n^*$ form the dual basis to the basis $e_1, \ldots, e_n$.
I claim that $\mathcal{C} = \bigcap_{i = d + 1}^n \ker e^*_i$. Note that if $1 \le j \le d$ and $d + 1 \le i \le n$, then $$e_i^*(e_j) = 0 \implies e_j \in \ker e_i^*,$$ hence $$\mathcal{C} = \operatorname{span}(e_1, \ldots, e_d) \subseteq \bigcap_{i=d+1}^n \ker e_i^*.$$ Conversely, suppose $x \in \bigcap_{i=d+1}^n e_i^*$. Since $x \in F_q^n$, we have $x = a_1 e_1 + \ldots + a_n e_n$ for some scalars $a_1, \ldots, a_n \in F_q$. We have, for $d + 1 \le i \le n$, $$0 = e_i^*(x) = a_1 e_i^*(e_1) + \ldots + a_{i - 1} e_i^*(e_{i - 1}) + a_i e_i^*(e_i) + a_{i + 1} e_i^*(e_{i + 1}) + \ldots + a_n e_i^*(e_n) = a_i,$$ hence $$x = a_1 e_1 + \ldots + a_d e_d + 0 + \ldots + 0 \in \mathcal{C}.$$ Thus, $\mathcal{C}$ can indeed be expressed as the intersection of the kernels of $n - d$ linear forms.