Let $E$ be a normed linear space and $G$ a $n$-dimensional subspace of $E$. Let $\{e_1, \ldots, e_n\}$ be a basis of $G$. Then every $x\in G$ can be written as $x= \sum_{i=1}^n x_i e_i$. Let $f_i: G \to \mathbb R, x \mapsto x_i$. Then $f_i$ is linear continuous. By Hahn-Banach theorem, each $f_i$ admits a linear continuous extension $\hat f_i: E \to \mathbb R$, i.e., $\hat f_i \restriction G = f_i$. Let $$L := \bigcap_{i=1}^n \operatorname{ker} \hat f_i.$$
It's easy to show that $L$ is a closed subspace of $E$ such that $G \cap L = \{0\}$. Could you leave some hints to show that $G+L =E$?
The inclusion $G + L \subseteq E$ is trivial.
For the other inclusion, let $x \in E$ and consider $y= \sum_{i=1}^n \hat{f}_i(x)e_i.$ Clearly $y \in G.$ Now show that $x-y\in L$ by checking that $\hat{f}_j(x-y)=0$ for all $j=1,\dots,n.$ Then you have $x = y + (x-y) \in G + L.$