Let $V$ be a normed space. Let $\gamma\colon [a,b] \rightarrow V$ be continuous. Then $\gamma$ is a curve.
Let $P$ be a partition of $[a,b]$, then $$\Lambda(\gamma, P) := \sum_{i=1}^n \| \gamma(x_i) - \gamma(x_{i-1})\|$$ We usually define the length of that curve as $$\Lambda(\gamma) := \sup_P \Lambda(\gamma, P)$$
A classical theorem of analysis is if $\|\cdot\|$ is an Euclidean norm and $\gamma \in C^1$, then: $$\Lambda(\gamma) = \int_a^b \| \gamma'(t)\| dt$$
Is the theorem true if $\|\cdot\|$ is not an Euclidean norm? The proof in Rudin's Principles of the Mathematical Analysis uses this at the start.
In Rudin's PMA (Theorem 6.27) $\gamma$ is assumed continuously differentiable. The proof given there works for general normed spaces with no changes: see below. However, it invokes the integral triangle inequality $$ \left\|\int_a^b \mathbf f(t)\,dt \right\| \le \int_a^b \left\|\mathbf f(t)\right\|\,dt \tag{1}$$ with $\mathbf f = \gamma'$. Since the proof of (1), Theorem 6.25, explicitly involves the Euclidean norm, a more general proof should be supplied. Here it is.
Let $I = \int_a^b \mathbf f(t)\,dt$. By the Hahn-Banach theorem there exists a unit-norm functional $\varphi\in V^*$ such that $\varphi(I)=\|I\|$. The function $\varphi\circ \mathbf f$ is continuous, as a composition of continuous functions. Also, $\varphi\circ \mathbf f \le \|\mathbf f\|$ because $\varphi$ has unit norm. Therefore, $$ \|I\| = \varphi (I) = \int_a^b \varphi(\mathbf f(t))\,dt \le \int_a^b \|\mathbf f(t)\|\,dt $$ as claimed.
For completeness, I sketch the proof of Theorem 6.27. The integral triangle inequality gives one half of it:
$$ \Lambda(\gamma, P) = \sum_i \left\|\int_{x_{i-1}}^{x_i} \gamma'(t)\,dt \right\| \le \sum_i \int_{x_{i-1}}^{x_i} \left\|\gamma'(t)\right\| \,dt = \int_{a}^{b} \left\|\gamma'(t)\right\| \,dt $$
The converse: by the mean value theorem, each difference $\|\gamma(x_i)-\gamma(x_{i-1})\|$ is equal to $\|\gamma'(t_i)\|(x_i-x_{i-1})$ for some $t_i\in (x_{i-1},x_i)$. This implies $L(P, \|\gamma'\|, dt)\le \Lambda(\gamma, P)\le \Lambda(\gamma)$. Since the integral $\int_{a}^{b} \left\|\gamma'(t)\right\| \,dt$ is the supremum of $L(P, \|\gamma'\|, dt)$ over $P$, it follows that $$\int_{a}^{b} \left\|\gamma'(t)\right\| \,dt \le \Lambda(\gamma)$$