Let $V$ be a finite-dimensional real vector space with its Euclidean topology. Then all norms on $V$ are equivalent and consequently given two norms $\lVert-\rVert$, $\lVert-\rVert'$, the number
$$
d = d(\lVert-\rVert, \lVert-\rVert') := \sup\_{0 \neq v \in V}\big| \log\lVert v\rVert - \log\lVert v\rVert'\big|
$$
is finite. This equips the set of all norms on $V$ with the structure of a complete metric space. You can think of $C := \exp(d)$ as the smallest real number $\ge 1$ such that $1/C \cdot \lVert-\rVert' \le \lVert-\rVert \le C \cdot \lVert-\rVert'$.


Now let $\{\lVert-\rVert\_t\}\_{t \in \mathbf{R}}$ be a family of norms on $V$, parametrized by the real numbers (I would also be interested in more general parameter spaces). Let us say that this is a *continuously varying family of norms* on $V$ if the map
$$
\mathbf{R} \times V \to \mathbf{R}\_{\ge 0},
\qquad
(t, v) \mapsto \lVert v\rVert\_t
$$
is continuous.


**Question:** Let $\{\lVert-\rVert\_t\}\_{t \in \mathbf{R}}$, $\{\lVert-\rVert'\_t\}\_{t \in \mathbf{R}}$ be two continuously varying families of norms on $V$. Will the function
$$
\mathbf{R} \to \mathbf{R}\_{\ge 0},
\qquad
t \mapsto d(\lVert-\rVert\_t, \lVert-\rVert'\_t)
$$
be continuous?



Continuously varying norms
I expand my comment where I claim that, on the space (call it $N(V)$) of all norms on $V$, the smallest topology making continuous the evaluations $\|\cdot\| \mapsto \|v\|$ (for $v \in V$) coincides with the topology defined by the distance $d$.


This clearly implies that, under your hypothesis, the maps $t \mapsto \|\cdot\|\_t$ and $t \mapsto \|\cdot\|'\_t$ are continuous for the distance $d$, and in particular your question has a positive answer. This holds more generally if $T$ is an arbitrary topological space, $\{\|\cdot\|\_t\}\_{t \in T}$ is a family indexed by $T$ of norms on $V$, and if for every $v \in V$, $t\mapsto \|v\|\_t$ is continuous.


Let me justify the initial claim. A first observation is that, by homogeneity, if $B$ denotes the closed unit ball for a fixed norm on $V$, the topology given by $d$ coincides with the topology of uniform convergence on $B$. Moreover, $V$ being finite dimensional, $B$ is compact and therefore the topology given by $d$ coincides with the topology of uniform convergence on compact subsets of $V$.


On the other hand let $v\_1,\dots,v\_n$ be a basis for $V$. Observe that, for an arbitrary $v=\sum\_i t\_i v\_i$, $v'=\sum\_i t'\_i v\_i$ and a norm $\|\cdot\|$ on $V$, we have
$$ \left|\|v\| - \|v'\| \right| \leq \|v-v'\| \leq \sum\_i |t\_i-t'\_i| \|v\_i\|.$$
This implies that, for every constant $C$, $\{ \|\cdot\| \in N(V) | \forall i\leq n, \|v\_i\| < C\}$ is made of equicontinuous functions on $V$. By the [the Arzelà-Ascoli Theorem](https://en.wikipedia.org/wiki/Arzel%C3%A0%E2%80%93Ascoli_theorem), in restriction to this set, the uniform convergence on compact subsets of $V$ coincides with pointwise convergence.


This proves the claim because the sets $\{ \|\cdot\| \in N(V) | \forall i\leq n, \|v\_i\| < C\}$ form a exhaustion of $N(V)$ by open sets.


