Your proof is correct. Well done!
As for the other result, here's one way to think about it:
Take $e_j$ to be the standard basis of $\Bbb R^n$, and take $f_j$ to be the corresponding dual basis. The system of equations
$$
a_{11} x_1 + \cdots + a_{1n} x_n = 0\\
a_{21} x_1 + \cdots + a_{2n} x_n = 0\\
\vdots \\
a_{m1} x_1 + \cdots + a_{mn} x_n = 0
$$
can be rewritten as
$$
(a_{11}f_1 + \cdots + a_{1n}f_n) x = 0\\
(a_{21}f_1 + \cdots + a_{2n}f_n) x = 0\\
\vdots\\
(a_{m1}f_1 + \cdots + a_{mn}f_n) x = 0
$$
That is, the solution set of the system of equation is the common zero set of the linearly independent set of functionals $\sum_{j=1}^n a_{ij}f_j$ where $i = 1,\dots,m$. That is, if $M$ denotes the solution set to the homogeneous system of equations, we have
$$
M^a = \text{span}\left\{\sum_{j=1}^n a_{1j}f_j,\dots,\sum_{j=1}^n a_{mj}f_j\right\}
$$
By your result, we may conclude $\dim(M) = n-m$, as desired.
After the fruitful discussion, the full proof goes as follows.
Since the normed space $X$ is finite dimensional, then every linear operator on $X$ is bounded (or Shortly $X'=X^*$ where $X^*$ is algebraic dual space of $X$) as stated in Theorem 2.7-8 in Erwin Kreyszig's book.
If $\{e_1,\cdots,e_n\}$ is a basis for $X$, then each $x\in X$ is written as
$x=\sum_{k=1}^{n} \xi_k e_k$
and because $f\in X'$ is linear
\begin{equation}
f(x)=f\left(\sum_{k=1}^{n} \xi_k e_k\right) = \sum_{k=1}^{n} \xi_k \underbrace{f(e_k)}_{=\gamma_k} = \sum_{k=1}^{n} \xi_k \gamma_k
\end{equation}
Next,
\begin{equation}
\left | f(x) \right |=\left | \sum \xi_k \gamma_k \right |\leq \sum \left | \xi_k \gamma_k \right | \leq \left \| x \right \| \sum \left | \gamma_k \right |
\end{equation}
and because $f\in X'$ is bounded,
$\left \| f \right \| = \sup_{\left \| x \right \|\neq 0} \frac{\left | f(x) \right |}{\left \| x \right \|}\leq \frac{\left \| x \right \|\sum \gamma _{k}}{\left \| x \right \|}=\sum \gamma _{k}$
which gives the 1st direction
$\left \| f \right \|\leq \sum \left | \gamma _{k} \right |$
on the other hand if we consider $x=\left ( \xi_k \right )$ with
$\xi_{k}=
\left\{\begin{matrix}
\left | \gamma_{k} \right | /\gamma_{k} & \text{if}\ \gamma _{k}\neq 0\\
0 & \text{if}\ \gamma _{k}= 0
\end{matrix}\right.$
we observe that $\left \| x \right \| = 1$ and we also observe that
$\left | f(x) \right |=\sum \left | \gamma _{k} \right |\leqslant \left \| f \right \| \left \| x \right \|=\left \| f \right \|$
which gives the second direction
$\sum \left | \gamma _{k} \right |\leq \left \| f \right \|$
comparing the 1st direction and the second direction, we conclude
$\left \| f \right \| = \sum \gamma _{k}$ which is the norm on $l^1$
Best Answer
What you have done is correct and you have shown $||f||_{c_0'} \le ||\hat f||_{\ell_1}$, where $\hat f$ is the sequence given by $\hat f_i = f(e_i)$. On the other hand, let $g_n\in c_0$ be defined so that
\begin{equation} (g_n)_i = \begin{cases} 1 & \text{ if } f(e_i) \ge 0, i\le n \\ -1 & \text{ if } f(e_i)<0 , i\le n\\ 0 & \text{ if } i >n \end{cases} \end{equation}
Then $||g_n||_{c_0} =1$ and
$$f(g_n) = \sum_{i=1}^n |f(e_i)|. $$
This implies
$$||f||_{c_0'} \ge \sum_{i=1}^n |f(e_i)|$$
for all $n\in \mathbb N$. Take $n\to \infty$ we have the opposite inequality. Thus
$$||f||_{c_0} = ||\hat f||_{\ell_1}.$$
Thus the map $c_0' \to \ell_1$, $f\mapsto \hat f$ is an isometric embedding. Also one can see that this is surjective, thus as a Banach space we have $c_0' \cong \ell_1$.