It looks like the "logic aspects" of the argument boil down to using compactness.

First, given a bound $f:C\to[0,\infty)$. The set $\mathcal{L}_f$ of all 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 norm in $\mathcal{L}_f$. Therefore, unless empty, $\mathcal{L}_f$.

Given a gapped norm $L$ (including the requirement $L(X \oplus Y) = L(X) + L(Y)$. Set $f_n(X) = \frac{1}{n}\lceil L(X^{\oplus n}) \rceil$. Since $L \leq f_n$, $\mathcal{L}_{f_n}$ is nonempty, so it contains a maximal element $L_n$. Note that $\lim_{n\to\infty} L_n = L$.

Claim 1: $L_n$ takes values in $\frac{1}{n}\mathbb{N}$.

Proof. Given a finite list of distinguished triangles $(A_i,B_i,C_i)$, $i < \ell$. Construct a sequence of functions $h_0 \geq h_1 \geq \cdots$ as follows: Start with $h_0 = f_n$. Once $h_k$ is defined, set $h_{k+1}(B_i) = \min(h_k(B_i),h_k(A_i) + h_k(C_i))$.

Each $h_k$ takes values in $\frac{1}{n}\mathbb{N}$. Since $\frac{1}{n}\mathbb{N}$ is discrete, the sequence must stabilize to a limit $h$. Moreover, $L_n \leq h$.

By compactness, there is an accumulation point to all such $h$'s along the net of all finite lists of distinguished triangles. That accumulation point must be a norm and therefore it must be exactly $L_n$, by maximality.

Claim 2: Given $X \in C$ and any prime $p$, there is a subsequence of $L_{p^i}$ along which $L_{p^i}(X)$ is constant.

[I don't follow all the details here but this appears to be shown in the proof outline. Of course, this claim is obvious from the conclusion since $L_n$ must in fact be a constant sequence.]

From claim 2, we see that $L(X) \in \mathbb{N}[\frac{1}{p}]$. Since $p$ is arbitrary, it follows that $L(X) \in \mathbb{N}$.

It looks like the "logic aspects" of the argument boil down to using compactness. [However, this appears to be moot since the argument may have flaws pointed out by Will Sawin.]

First, given a bound $f:C\to[0,\infty)$. The set $\mathcal{L}_f$ of all 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 norm in $\mathcal{L}_f$. Therefore, unless empty, $\mathcal{L}_f$ has a maximal element.

Given a gapped norm $L$ (including the requirement $L(X \oplus Y) = L(X) + L(Y)$. Set $f_n(X) = \frac{1}{n}\lceil L(X^{\oplus n}) \rceil$. Since $L \leq f_n$, $\mathcal{L}_{f_n}$ is nonempty, so it contains a maximal element $L_n$. Note that $\lim_{n\to\infty} L_n = L$.

Claim 1: $L_n$ takes values in $\frac{1}{n}\mathbb{N}$.

Proof. Given a finite list of distinguished triangles $(A_i,B_i,C_i)$, $i < \ell$. Construct a sequence of functions $h_0 \geq h_1 \geq \cdots$ as follows: Start with $h_0 = f_n$. Once $h_k$ is defined, set $h_{k+1}(B_i) = \min(h_k(B_i),h_k(A_i) + h_k(C_i))$.

Each $h_k$ takes values in $\frac{1}{n}\mathbb{N}$. Since $\frac{1}{n}\mathbb{N}$ is discrete, the sequence must stabilize to a limit $h$. Moreover, $L_n \leq h$.

By compactness, there is an accumulation point to all such $h$'s along the net of all finite lists of distinguished triangles. That accumulation point must be a norm and therefore it must be exactly $L_n$, by maximality.

Claim 2: Given $X \in C$ and any prime $p$, there is a subsequence of $L_{p^i}$ along which $L_{p^i}(X)$ is constant.

[I don't follow all the details here but this appears to be shown in the proof outline. Of course, this claim is obvious from the conclusion since $L_n$ must in fact be a constant sequence.]

From claim 2, we see that $L(X) \in \mathbb{N}[\frac{1}{p}]$. Since $p$ is arbitrary, it follows that $L(X) \in \mathbb{N}$.

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.

It looks like the "logic aspects" of the argument boil down to using compactness.

First, given a bound $f:C\to[0,\infty)$. The set $\mathcal{L}_f$ of all 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 norm in $\mathcal{L}_f$. Therefore, unless empty, $\mathcal{L}_f$.

Given a gapped norm $L$ (including the requirement $L(X \oplus Y) = L(X) + L(Y)$. Set $f_n(X) = \frac{1}{n}\lceil L(X^{\oplus n}) \rceil$. Since $L \leq f_n$, $\mathcal{L}_{f_n}$ is nonempty, so it contains a maximal element $L_n$. Note that $\lim_{n\to\infty} L_n = L$.

Claim 1: $L_n$ takes values in $\frac{1}{n}\mathbb{N}$.

Proof. Given a finite list of distinguished triangles $(A_i,B_i,C_i)$, $i < \ell$. Construct a sequence of functions $h_0 \geq h_1 \geq \cdots$ as follows: Start with $h_0 = f_n$. Once $h_k$ is defined, set $h_{k+1}(B_i) = \min(h_k(B_i),h_k(A_i) + h_k(C_i))$.

Each $h_k$ takes values in $\frac{1}{n}\mathbb{N}$. Since $\frac{1}{n}\mathbb{N}$ is discrete, the sequence must stabilize to a limit $h$. Moreover, $L_n \leq h$.

By compactness, there is an accumulation point to all such $h$'s along the net of all finite lists of distinguished triangles. That accumulation point must be a norm and therefore it must be exactly $L_n$, by maximality.

Claim 2: Given $X \in C$ and any prime $p$, there is a subsequence of $L_{p^i}$ along which $L_{p^i}(X)$ is constant.

[I don't follow all the details here but this appears to be shown in the proof outline. Of course, this claim is obvious from the conclusion since $L_n$ must in fact be a constant sequence.]

From claim 2, we see that $L(X) \in \mathbb{N}[\frac{1}{p}]$. Since $p$ is arbitrary, it follows that $L(X) \in \mathbb{N}$.