In the book Functional Analysis of Rudin. The theorem 1.28 says that
(a) If $d$ is a translation-invariant metric on a vector space $X$ then $d(nx,0)\leq nd(x,0)$ for every $x\in X$ and for $n=1,2,3\cdots$
(b) If $\{x_n\}$ is a sequence in a metrizable topological vector space $X$ and if $x_n\to 0$ as $n\to\infty$, then there are positive scalars $\gamma_n$ such that $\gamma_n\to\infty$ and $\gamma_mx_n\to 0$
In the book it says
To prove (b), let $d$ be a metric as in (a),...
Why it can choose a translation-invariant metric ?
I know that in this book theorem 1.24 says that if $X$ is a topological vector space with a countable local base, then there is a metric $d$ on $X$ such that $d$ is compatible with the topology of $X$ and $d$ is invariant.
But theorem 1.28 doesn't have the hypothesis. Does the conclusion still hold?
If $X$ is a metrizable topological space (you don't even need the linear structure here), then every point has a countable neighborhood base. For example, if $d$ is a metric that induces the topology of $X$ and if $U_r(x)$ denotes the open ball of radius $r$ around $x$ (with respect to $d$), then $\{B_{1/n}(x)\mid n\in\mathbb N\}$ is a countable neighborhood base at $x$. So in your case, you can apply Theorem 1.24 since $X$ is assumed to be a metrizable topological vector space.