I will use the following definition of "finite dimensional":
Definition A vector space $V$ is finite dimensional if $V$ has a finite spanning set. Otherwise, $V$ is infinite dimensional.
Note: the definition does not presume that we know what "dimension" means.
One has the following elementary theorem about matrices:
Theorem 1 If a matrix $A$ has more columns than rows, then there exists an non-zero solution to $A x = 0$.
Proof: Row reduction.
The theorem has the following consequence:
Theorem 2 Suppose $V$ is finite dimensional with a spanning set of cardinality $n$. Then any linearly independent set in $V$ has cardinality $\le n$.
Proof. Suppose $\{v_1, \dots, v_n\}$ is a spanning set and $\{w_1,
\dots, w_m\}$ is linearly independent. Since $\{v_1, \dots, v_n\}$ is spanning, there exist $a_{i, j}$ in the underlying field $K$ such that
$w_j = \sum a_{i, j} v_i$. Let $A$ denote the matrix $(a_{i, j})$. If $m >n$, there exists $$x = \begin{bmatrix}x_1 \\ \vdots \\ x_m \end{bmatrix} \ne 0$$ such that $A x = 0$. Check that this implies $\sum_j x_j w_j = 0$, contradicting the linear independence of $\{w_1,
\dots, w_m\}$.
With a tiny bit more effort, one can prove the existence of bases and that any two bases have the same cardinality:
Definition A basis of $V$ is a set that is both spanning and linearly independent.
Theorem 3 A finite dimensional vector space has a basis.
Sketch: Choose a spanning set of minimal cardinality and show that it is also linearly independent.
Theorem 4 Any two bases of a finite dimensional vector space have the same cardinality.
Proof. Follows from Theorem 2.
Definition The cardinality of a basis is called the dimension of $V$.
Theorem Let $V$ be a finite dimensional vector space with dimension $n$. Then any spanning set of $V$ has cardinality $\ge n$ and any linearly independent subset of $V$ has cardinality $\le n$.
Proof. Follows from Theorem 2.