The exercise asks to prove that for a given neighbourhood of $0$, $V$, in a topological vector space $X$, there exists a function $f:X\to\mathbb{R}$ such that $f(0)=0$ and $f(x)=1$, for any $x\notin V$. A suggestion is also given: Let $V_n$ be balanced neighbourhoods of $0$ such that $V_1+V_1\subseteq V$ and $V_{n+1}+V_{n+1}\subseteq V_n$. Construct $f$ as in the proof of Theorem 1.24. Show that $f$ is continuous at $0$ and that $|f(x)-f(y)|\leq f(x-y)$.
The proof of Theorem 1.24 is based on the following machinery: Let $$D=\{r\in\mathbb{Q}\cap[0,1)/ \ \exists k_r\in\mathbb{N}: \ r=\sum_{j=1}^{k_r} c_j(r)2^{-j}, \ c_j(r)=0,1\}.$$ Then define $$A(r)=X, \ \forall r\geq1, \ \text{and } A(r)=\sum_{j=1}^{k_r}c_j(r)V_j, \ \forall r\in D$$ and $$f(x)=\inf\{r\in\mathbb{Q}\cap[0,1]: \ x\in A(r)\}.$$
My question is about proving $|f(x)-f(y)|\leq f(x-y)$. As a matter of fact, this can be easily obtained if $f$ is subadditive. In the proof of Theorem 1.24 this is based on the fact that $A(r)+A(s)\subseteq A(r+s)$, for any $r,s\in D$. But the latter is proved by using the slightly different initial assumption: $V_{n+1}+V_{n+1}+V_{n+1}+V_{n+1}\subseteq V_n$ and based on an argument about addition in the binary system notation. I mean, of course, the latter condition also implies that $V_{n+1}+V_{n+1}\subseteq V_n$, so, the original proof still works, but I was just curious if we can prove that $|f(x)-f(y)|\leq f(x-y)$ by only using $V_{n+1}+V_{n+1}\subseteq V_n$.