I would like to clarify if the above statement is true.
Suppose $V$ is a finitely generated. Then every submodule of $V$ is a vector subspace of $V$ hence finitely generated so $V$ is Noetherian.
As for the Artinian portion: Suppose we have a sequence of decreasing submodules (i.e vector subspaces) $M_1\supset M_2 \supset....$, then $dim(M_1)\supset dim(M_2) \supset....$ is a decreasing sequence of non-negative integers so $\exists k$ such that $dim(M_k)=dim(M_{k+1})=...$ which implies that $M_k = M_{k+1}=...$.
Are the 2 proofs valid?
One other way to see it is to use the fact that over a Noetherian ring $R,$ an $R$-module $M$ is Noetherian if and only if it is finitely generated over $R,$ i.e., $M = R \langle m_1, \dots, m_n \rangle$ for some elements $m_i$ of $M.$ Likewise, if $R$ is Artinian, then we have that $M$ is Artinian if and only if it is finitely generated over $R.$
Claim. Every finite-dimensional vector space is Noetherian and Artinian.
Proof. Consider a finite-dimensional $k$-vector space $V.$ By hypothesis that $\dim_k V = n < \infty,$ there exist vectors $v_1, \dots, v_n$ in $V$ such that every vector $v$ in $V$ can be expressed as $v = a_1 v_1 + \cdots + a_n v_n$ for some elements $a_i$ in $k.$ We have therefore that $V = k \langle v_1, \dots, v_n \rangle$ so that $V$ is finitely generated over $k.$ Considering that $k$ is a field, it follows that $k$ is Noetherian and Artinian, hence $V$ is Noetherian and Artinian.