Direct proof of $\exists\ C > 0$ s.t. $\forall\ v \in V$, $\lVert v\rVert_1 \leq C\lVert v\rVert$

Problem

Let $V$ be a finite dimensional vector space with norm $\lVert \cdot \rVert$. Let $d(x,y) = \lVert x-y \rVert$ be the metric on $V$. Let $\{e_1, e_2, \ldots, e_n\}$ be a basis of $V$. If $v = \sum_{i = 1}^n v_ie_i$, then let $\lVert v \rVert_1 = \sum_{i = 1}^n |v_i|$. Then

$(1)$ $\exists\ C > 0$ s.t. $\forall\ v \in V$, $\lVert v\rVert_1 \leq C\lVert v\rVert$.

I want a direct way of proving $(1)$ without first proving $(2)$. Is that possible without an explicit formula for $\lVert \cdot \rVert$?

Background

I want to use $(1)$ to show that every linear transformation on a finite dimensional normed space is continuous as in Wikipedia. Wikipedia justified the use of $(1)$ by quoting

$(2)$ All norms on a finite dimensional space are equivalent.

Then I learnt how to prove $(2)$ from this first result from google. The doc essentially uses the metric $d'(x,y) = \lVert x-y \rVert_1$. Then it proves $(2)$ by first proving

$(3)$ $\lVert \cdot \rVert$ is continuous on $V$ with respect to metric $d'$.

, invoke extreme value theorem on the unit sphere, then uses the fact that norm equivalence is in fact an equivalent relation to show that norms are equivalent. This is a way to prove $(1)$.

Attempt

It suffices to show

$(4)$ $\lVert \cdot \rVert_1$ is continuous at $0_V$ with respect to metric $d$.

Then we can use the extreme value theorem on the unit sphere again. So that I also accept answer that proves $(4)$ without first proving $(2)$. Here is my failed attempt. Pick any $\epsilon > 0$. Let $m = \min_{1\leq i\leq n} \lVert e_i \rVert$.

$$\lVert v\rVert_1-\lVert 0_V\rVert_1 = \sum_{i=1}^n |v_i| = \frac{\sum_{i=1}^n m|v_i| }{m} \leq \frac{\sum_{i=1}^n |v_i| \lVert e_i \rVert}{m}=\frac{\sum_{i=1}^n \lVert v_ie_i \rVert}{m}$$

Taking $\delta = m\epsilon$ won't work since it is not true that $\sum_{i=1}^n \lVert v_ie_i\rVert \leq \lVert v\rVert$...