Does this compactness argument work?
First, given a bound $f:C\to[0,\infty)$. The set $\mathcal{L}_f$ of all gapped norms $L \leq f$ is a closed (and hence compact) subset of $\prod_{X \in C} [0,f(X)]$ because each inequality is a closed requirement.
Note that if $L_1, L_2 \in \mathcal{L}_f$ then the pointwise maximum $L(X) = \max(L_1(X),L_2(X))$ is also a gapped norm in $\mathcal{L}_f$. Therefore, unless empty, $\mathcal{L}_f$ has a maximal element $L_\infty$.
Given a gapped norm $L$. Set $f(X) = \lceil L(X) \rceil$. Then $\mathcal{L}_f$ is nonempty, so the maximal element $L_\infty$ exists and $L_\infty \geq L$.
We claim that $L_\infty = f$! To see this, it suffices to show that for every $X_0 \in C$ there is an $L_0 \in \mathcal{L}_f$ with $L_0(X_0) = f(X_0)$. In fact, it is enough to show that for any finite list $I_1,\ldots,I_k$ of gapped norm inequalities, there is a function $L_0 \in \prod_{X \in C} [0,f(X)]$ that meets the inequalities $I_1,\ldots,I_k$. [Fill in proof here...]
We claim that if moreover $L(X \oplus Y) = L(X) + L(Y)$ for all $X, Y \in C$, then $L_\infty = L$. This is because $L(X) = \lim_{n\to\infty} \frac{1}{n}\lceil L(X^{\oplus n}) \rceil$ and [fill in details here...]
It follows that $L$ must be integer valued.