Here is one approach, mentioned in Lang's analysis book.
Let $ V $ be an $ \mathbb{R} $-vector space with basis $ e_1 , \ldots, e_k $.
$ \| x_1 e_1 + \ldots + x_k e_k \|_\infty := \max\{ |x_1|, \ldots, |x_k| \} $ is a norm on $ V $.
$ \| \ldots \|_{\infty} $ makes $ V $ complete.
Let $ (v_j)_{j \geq 1} = (v_1, v_2, \ldots ) $ be a Cauchy sequence in $ V $. Exressing w.r.t basis vectors, $$ (v_1, v_2, \ldots ) = (e_1, e_2, \ldots, e_k) \begin{pmatrix} v_{11} &v_{12} &\ldots \\ v_{21} &v_{22} &\ldots \\ \vdots &\vdots &\ddots \\ v_{k1} &v_{k2} &\ldots \end{pmatrix}, \, \text{each }v_{ij} \in \mathbb{R} .$$ That is, $ v_{ij} $ is $ e_i $-coordinate of $ v_j $.
Notice sequence $ (v_{1j})_{j \geq 1} = (v_{11}, v_{12}, v_{13}, \ldots ) $ is Cauchy [because $ | v_{1n} - v_{1m} | \leq \| v_n - v_m \|_{\infty} $]. Similarly sequences $ (v_{1j}), (v_{2j}) , \ldots, (v_{kj}) $ are all Cauchy, hence convergent.
So let $ v_{1j} \to c_1, \ldots, v_{kj} \to c_k $ as $ j \to \infty $. Now $ v_j \to c_1 e_1 +\ldots + c_k e_k $ as $ j \to \infty $ [because $\| v_j - (c_1 e_1 + \ldots + c_k e_k) \|_\infty $ $ =\max \{ |v_{1j} - c_1| , \ldots, |v_{kj} - c_k | \} $ $ \to 0 $ as $ j \to \infty $], as needed.
Now let $ \| \ldots \| $ be any norm on $ V $. We'll try to show $ \| \ldots \| $ and $ \| \ldots \|_{\infty} $ are equivalent on $ V $, by induction on dimension $ k $ of the normed space. So lets assume $ \| \ldots \| $ and $ \| \ldots \|_{\infty} $ are equivalent on every proper subspace of $ V $.
Firstly, $$ \begin{align} \| x \| &= \| x_1 e_1 + \ldots + x_k e_k \| \\ &\leq |x_1| \| e_1 \| + \ldots + |x_k| \| e_k \| \\ &\leq \max \{|x_1 |, \ldots, |x_k|\} \left( \| e_1 \| + \ldots + \| e_k \| \right) \\ &= \| x \|_\infty C \end{align}$$ with $ C > 0 $.
So we need only show $ \| x \|_\infty \leq D \| x \| $ for some $ D > 0 $.
Say there is no such $ D $. Then for every integer $ n > 0 $, there is a $ v_n $ with $ \| v_n \|_\infty > n \| v_n \| $.
Notice $ v_n \neq 0 $, i.e. each $ v_n $ has some non-zero coordinate.
So on defining $ w_n $ to be $ v_n $ divided by the coordinate of $ v_n $ with maximum absolute value, we have $ \| w_n \|_\infty = 1 $ and $ \| w_n \| = \| v_n \|_\infty ^{-1} \| v_n \| $ $\big($and hence $ \| w_n \| < \frac{1}{n} \big) $. Also each $ w_n $ has $ 1 $ as a coordinate.
Expressing sequence $ (w_n)_{n \geq 1} $ w.r.t basis vectors, $$ (w_1, w_2, \ldots ) = (e_1, e_2, \ldots, e_k) \begin{pmatrix} w_{11} &w_{12} &\ldots \\ w_{21} &w_{22} &\ldots \\ \vdots &\vdots &\ddots \\ w_{k1} &w_{k2} &\ldots \end{pmatrix}. $$ Every column has a $ 1 $, and every entry is $ \leq 1 $ in magnitude.
So there is a $ T \in \{ 1, 2, \ldots, k \} $ such that row $ w_{T 1}, w_{T 2}, \ldots $ has infinitely many $ 1 $s. Let $ J := \{ j : w_{T j} = 1 \} $ be the positions at which the $1$s occur in this row.
We'll now focus on sequence $ (w_j - e_T)_{j \in J} $.
Firstly, as $ (w_n)_{n \geq 1} $ itself converges to $ 0 $ w.r.t $ \| \ldots \| $, this sequence $ (w_j - e_T)_{j \in J} $ converges to $ (-e_T) $ w.r.t $ \| \ldots \| $.
Note the sequence $ (w_j - e_T)_{j \in J} $ lies in the subspace $ V_T := \{ x_1 e_1 + \ldots + x_k e_k : x_T = 0 \} $. It is also Cauchy in $ V_T $ w.r.t. $ \| \ldots \| $, because $ \| (w_n - e_T)-(w_m - e_T) \| = \| w_n - w_m\| $ $\leq \| w_n \| + \| w_m \| < \frac{1}{n} + \frac{1}{m}$.
But $ V_T $ is complete w.r.t $ \| \ldots \|_{\infty} $, and $ \| \ldots \| $ is equivalent to $ \| \ldots \|_{\infty} $ on $ V_T $. So $ V_T $ is complete w.r.t $ \| \ldots \| $. Hence $ (w_j - e_T)_{j \in J} $ must converge to a point in $ V_T $ w.r.t $ \| \ldots \| $, that is $ (-e_T) \in V_T $, a contradiction.
So there indeed is a $ D > 0 $ such that $ \| x \|_\infty \leq D \| x \| $ in $ V $, as needed.