Intersection of kernels and linear dependence of functionals

Let $K$ denote the scalar field. Consider $F\colon X \to K^n$ given by

$$F(x) = \begin{pmatrix}L_1(x)\\ L_2(x)\\ \vdots \\ L_n(x)\end{pmatrix}.$$

Let $R = \operatorname{im} F \subset K^n$. We have an induced isomorphism $$\tilde{F}\colon X/\ker F \xrightarrow{\sim} R.$$

Since $\bigcap\limits_{k=1}^n \ker L_k = \ker F \subset \ker L$, we have an induced linear form $\tilde{L} \colon X/\ker F \to K$, and can pull that back to $R$ as $\hat{L} := \tilde{L} \circ \tilde{F}^{-1}$. We can extend $\hat{L}$ to all of $K^n$ (extend a basis of $R$ to a basis of $K^n$, and choose arbitrary values, e.g. $0$, on the basis vectors not in $R$). Thus there is a linear form $\lambda \colon K^n \to K$ with

$$\lambda \circ F = \lambda\lvert_R \circ F = \hat{L}\circ F = \tilde{L}\circ \tilde{F}^{-1}\circ F = \tilde{L} \circ \pi = L,$$

where $\pi \colon X \to X/\ker F$ is the canonical projection.

But every linear form $K^n\to K$ can be written as a linear combination of the component projections, so there are $c_1,\dotsc, c_n$ with

$$\lambda\begin{pmatrix}u_1\\u_2 \\ \vdots \\ u_n \end{pmatrix} = \sum_{k=1}^n c_k\cdot u_k,$$

and that means

$$L(x) = \lambda(F(x)) = \sum_{k=1}^n c_k\cdot L_k(x)$$

for all $x\in X$, or

$$L = \sum_{k=1}^n c_k\cdot L_k.$$

I have a proof in the case where $X$ is reflexive.

Suppose that $L$ is not a linear combination of the $L_i$'s. Let $C = \{\sum_{i=1}^n t_i L_i : t_i \in \mathbb{R}\} \subseteq X^*$. Then $C$ is a closed convex subset and $C \cap \{L\} = \emptyset$ by assumption. So by geometric Hahn-Banach, there exists $\xi \in X^{**}$ such that $\xi(C) \subseteq (-\infty,\alpha)$ and $\xi(L) > \alpha$. Since $C$ is a subspace, we actually have $\xi(C) = \{0\}$ and $\alpha > 0$. Assuming that $X$ is reflexive, then $\xi$ corresponds to evaluation at some $x \in X$. This shows that $L_i(x) = 0$ for all $i = 1, \ldots, n$ but $L(x) > 0$. This contradicts $\bigcap_{i=1}^n \ker L_i \subseteq \ker L$.

I'm not sure whether this extends to the case where $X$ is not reflexive, however.