This answer is somewhat verbose, it might helpt to just draw the corresponding picture for yourself.
We are assuming that $V=\mathbb{Q}^d$ with the subspace topology inherited from the Euclidean topology on $\mathbb{R}^d$ (the OP clarified in the comments that this is the setting of interest to them). Furthermore, we will assume that $d\geq 2$ (the case $d=1$ is trivial as $C$ would be a half-space and $f(x) = \vert x\vert^m f(e)$ for $x\in C^\mathrm{int}$ and $e=1$ if $C=[0,\infty)$ and $e=-1$ if $C=(-\infty,0]$. Clearly $f$ is locally uniformly continuous). In this answer, we denote by $B(x,r)=\{v\in \mathbb{Q}^d \ : \ \vert v-x\vert <r\}$ the ball in $\mathbb{Q}^d$ of radius $r$ centered at $x$.
We want to prove that the map $f$ is locally uniformly bounded on $C^\mathrm{int}$. The only thing which is of quantitative nature in our assumption is the requirement that
$$ f(c x) = c^m f(x) $$
for all $c\in \mathbb{Q}_{>0}$ and all $x\in C$. We would like to make use of this to estimate the difference between close points. Namely, if we want to compare $x,y\in C^\mathrm{int}$, then we would like to make a "zigzag" between $x$ and the ray spanned by $y$. More precisely, we would like to find $\varepsilon(x,y)>0$ such that
$$ y-(1-\varepsilon(x,y))x, (1+\varepsilon(x,y))x-y\in C^\mathrm{int}. $$
In that case we have by conditions $1$ and $2$ that
$$ \left( 1-\varepsilon(x,y) \right)^m f(x) = f((1-\varepsilon(x,y)x) \leq f(y) \leq f((1+\varepsilon(x,y))x) = \left( 1+\varepsilon(x,y) \right)^m f(x). $$
In particular, this implies
$$ \vert f(x)-f(y) \vert \leq \left[ \left( 1+\varepsilon(x,y) \right)^m - \left( 1-\varepsilon(x,y) \right)^m \right] f(x). $$
We need to make sure that both $\varepsilon(x,y)$ are locally uniformly controlled and $f(x)$ too.
Let us fix some $e\in C^\mathrm{int}$ and some $0<\delta <\vert e\vert/8$ such that $B(e,4\delta)\subset C^\mathrm{int}$. In particular, this means that for every $x\in B(e,\delta)$ the cone $C_x = \mathbb{Q}_{>0} B(x,2\delta)$ generated by $B(x,2\delta)$ is contained in $C^\mathrm{int}$. There exists $\theta \in (0,\pi/2)$ such that all $C_x$ contain a cone of the form
$$ \{ v\in \mathbb{Q}^d \ : \ 0<\langle v,x \rangle \leq\cos(\theta) \vert x \vert \} \subseteq C_x \subseteq C^\mathrm{int}.$$
In particular, this means that if we were in $\mathbb{R}^d$ (instead of $\mathbb{Q}^d$), then we could just use (highschool) geometry to compute $\varepsilon(x,y)$ for $y\in B(x,\delta)$. Let's do this first.
We can rotate everything to reduce to the case where $x=(\vert x \vert, 0, \dots, 0)\in \mathbb{R}^d$ and $y=(y_1, y_2, 0, \dots, 0)\in \mathbb{R}^d$ and $y_1, y_2\geq 0$ (so really we are in $\mathbb{R}^2$). We are asking where we need to start on the ray generated by $(1,0, \dots, 0)$ to reach $y$ by adding only a vector in $C_{(\vert x \vert,0,\dots, 0)}$. The cone $C_{(\vert x \vert,0,\dots, 0)}$ has an opening angle of $\theta$ around $\mathbb{R}_{\geq 0}(1, 0, \dots, 0)$, thus, starting at $( y_1-y_2 \cot(\theta), 0, \dots, 0)$ would do the job. Similarly, if we start at $y$ and need to reach the ray generated by $(1,0, \dots, 0)$ by adding only a vector in $C_{(\vert x \vert,0,\dots, 0)}$, we can choose
$$ (y_1+y_2 \cot(\theta), 0, \dots, 0). $$
We have
$$ \vert \vert x \vert - (y_1\pm\cot(\theta)y_2)\vert \leq \vert x-y\vert (1+\cot(\theta)).$$
Here we have used that $0\leq y_1\leq \vert x-y\vert$.
All of those computations assumed that we are in $\mathbb{R}^d$. In order to account for the fact that not all those numbers would be in the reals, we just introduce an additional factor of $2$ (if we'd really care about constants, we could drop it by just approximating suitably and noting that everything is well-behaved). Then we one sees that
$$ \varepsilon(x,y) = 2 \vert x-y\vert (1+\cot(\theta)) $$
is an admissible choice (recall that $\theta$ is constant in the choosen neighborhood).
So far we have shown that there exists $\theta\in (0, \pi/4)$ such that for all $x,y\in B(e,2\delta)$ we get
$$ \vert f(x)-f(y) \vert \leq \left( \left[ 1+2 \vert x-y\vert (1+\cot(\theta)) \right]^m - \left[ 1-2 \vert x-y\vert (1+\cot(\theta)) \right]^m \right) f(x). $$
Using the same argument as before, comparing $x$ to $e$, we get for all $x,y\in B(e,2\delta)$
$$ \vert f(x)-f(y) \vert \leq \left( \left[ 1+2 \vert x-y\vert (1+\cot(\theta)) \right]^m - \left[ 1-2 \vert x-y\vert (1+\cot(\theta)) \right]^m \right) \left( 1+ 2\delta(1+\cot(\theta)) \right) f(e). $$
This proves that $f$ is locally uniformly continuous on $C^\mathrm{int}$ as there exists a constant $C_{\theta,m}$ depending only on $\theta,m$ such that for all $x,y\in \mathbb{R}^d$ holds
$$ \left[ 1+2 \vert x-y\vert (1+\cot(\theta)) \right]^m - \left[ 1-2 \vert x-y\vert (1+\cot(\theta)) \right]^m \leq C_{\theta,m} \vert x-y\vert.$$
Note that this proof gives us a slightly stronger result. Namely, we get that $f$ is locally lipschitz on $C^\mathrm{int}$.
