= Solution
A functional $p:X\to\mathbb R$ is <sublinear functional>[sublinear] when
$$
p(u+v)\leq p(u)+p(v),\qquad p(tu)=tp(u)\quad(t\geq0).
$$
Any extension to $\widetilde M=M+\mathbb Rx$ must have the form
$$
\widetilde\ell(y+tx)=\ell(y)+tc.
$$
Choose $c$ satisfying
$$
\sup_{y\in M}\{\ell(y)-p(y-x)\}
\leq c\leq
\inf_{y\in M}\{p(y+x)-\ell(y)\}.
$$
This interval is nonempty: for $y,z\in M$,
$$
\ell(y)+\ell(z)=\ell(y+z)
\leq p(y+z)
\leq p(y-x)+p(z+x),
$$
which rearranges to the required lower bound being at most the upper bound. For $t>0$, positive homogeneity reduces
$\widetilde\ell(y+tx)\leq p(y+tx)$ to the upper inequality after replacing $y$ by $y/t$; for $t<0$ it reduces to the lower inequality. The case $t=0$ is the hypothesis on $\ell$. Thus this choice gives the required dominated linear extension.
The dominated form of the <Hahn-Banach theorem> states that a linear functional on a subspace, bounded above by a sublinear functional, extends linearly to the whole real vector space while retaining that bound.
Let $M=\operatorname{span}\{z_1,\ldots,z_n\}$. Its coordinate maps
$$
\lambda_j\!\left(\sum_ka_kz_k\right)=a_j
$$
are continuous because every linear map on a finite-dimensional normed space is continuous. Hahn--Banach extends each to
$\ell_j\in Z'$ without increasing its norm, and then
$$
\ell_j(z_k)=\delta_{jk}.
$$
For an arbitrary finite-dimensional subspace $M$, choose a basis $z_1,\ldots,z_n$ and these extended coordinate functionals. Then
$$
N=\bigcap_{j=1}^n\ker\ell_j
$$
is closed. Every $z\in Z$ has the decomposition
$$
z=\sum_j\ell_j(z)z_j+
\left(z-\sum_j\ell_j(z)z_j\right)\in M+N.
$$
Applying every $\ell_j$ shows $M\cap N=\{0\}$, so
$$
\boxed{Z=M\oplus N}.
$$
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