Let $(E,F_1)$ be a Finsler vector bundle over a manifold $M$. (See precise definition below). Let $F_2$ be another Finsler function (norm) on $E$.
For any $p \in M \, , \, F_1|_{E_p}:E_p \to \mathbb{R}$ is a norm on a finite dimensional vector space. Hence the corresponding unit sphere $\mathbb{S^p_1}=\{v_p \in E_p |F_1(v_p)=1 \}$ is compact. So $F_2$ attains a minimium on $\mathbb{S^p_1}$. Thus we obtain a function $m:M \to \mathbb{R}$ via $m(p) = \min\{F_2(v_p) | v_p \in \mathbb{S^p_1}\}$
Question:
Is $m$ always continuous?
Remarks:
(1) I am quite sure $m$ is not smooth in general. For example, if $E=TM$ and $F_i$ are the Finsler norms induces by two Riemannian metric $g_1,g_2$, then a calculation here shows that $m(p)=\min \lambda(G)$ where $G$ is the component matrix $g_{ij}$ of one metric w.r.t an orthonormal frame of the other.
(2) We cannot always choose a continuous minimizing section $s:M \to E$ ,i.e $s$ such that: $s_p \in \mathbb{S^p_1} \, , \, F_2(s_p)=m(p)$
(This follows from the above example together with this answer)
Of course, when such a continuous choice as described in (2) is possible this imediately implies continuity of $m$.
A Finsler vector bundle is a (smooth) vector bundle $E$ over a (smooth) manifold $M$ together with a Finsler function $F : E \to \mathbb{R}$ such that for every vector $v \in E$:
(1) $F$ is smooth on the complement of the zero section of $E$.
(2) $F(v) \ge 0$ with equality if and only if $v = 0$ (positive definiteness).
(3) $F(\lambda v) = |\lambda| F(v)$ for all $\lambda \in \mathbb{R}$ (homogeneity).
(4) $F(v + w) \le F(v) + F(w)$ for every $w$ which is in the same fiber with v (subadditivity).
Answer: Yes, $m$ is always continuous.
Remark: Your definition of a Finsler function is contradictory. In part (3) you should either assume $F(\lambda v) = |\lambda| F(v)$ or restrict to $\lambda\geq0$. Some authors do not require Finsler functions to be symmetric.
Details: Suppose $m$ is not continuous. Then there is $p\in M$ and a sequence $(p_i)$ converging to $p$ so that $\lim_{i\to\infty}m(p_i)\neq m(p)$. A priori (from considerations of continuous of functions on metric spaces alone), the limit $\ell:=\lim_{i\to\infty}m(p_i)$ can be anything on $[0,\infty]$.
For each $i$, let us pick $v^i\in E_{p_i}$ so that $F_1(v^i)=1$ and $F_2(v^i)=m(p_i)$. We can identify the tangent spaces near $p$ (this comes from the very definition of a bundle) and pass to a subsequence to assume that the vectors $v^i$ converge to a limit $v\in E_p$. Since $F_1$ and $F_2$ depend continuously on the base point, we have $F_1(v)=1$ and $F_2(v)=\ell\in(0,\infty)$. By the definition of $m$ we must have $\ell\geq m(p)$. Since we have assumed $\ell\neq m(p)$ (to the end of finding a contradiction), we can conclude that $m(p)<\ell<\infty$.
Let $u\in E_p$ be a vector with $F_1(u)=1$ and $F_2(u)=m(p)$. With the aforementioned identification we can consider $u=u^i$ as a vector on $E_{p_i}$ as well. It follows from the continuous dependence of $F_1$ and $F_2$ on the base point that $\lim_{i\to\infty}F_1(u^i)=1$ and $\lim_{i\to\infty}F_2(u^i)=m(p)$. But $F_2(u^i)\geq F_1(u^i)m(p_i)$, so $$ m(p) = \lim_{i\to\infty}F_2(u^i) \geq \lim_{i\to\infty}F_1(u^i) \times \lim_{i\to\infty}m(p_i) = 1\times\ell > m(p). $$ This is impossible.
(This proof could have been organized differently, of course. I hope this conveys the key ideas clearly enough.)