Let $x,y \in \Bbb{Q}^×$, and for $\varepsilon＞0$, suppose $|x-y|_p＜\varepsilon$, where $| \ \ |_p$ is the $p$-adic absolute value.
My book says that if $\varepsilon$ is small enough, then $|y|_p＞(1/2)|x|_p$.
I think this seems obvious because $|y|_p$ and $|x|_p$ are close enough, so half of $|x|_p$ is smaller than $|y|_p$. But for what $t > 0$ does the condition that $\varepsilon ＜t$ imply $|y|_p ＞ (1/2)|x|_p$ when $|x-y|_p < \varepsilon$? I want to know what can be used for $t$.
Here I use $|\cdot|$ as the usual p-adic absolute value. There is no $t>0$ such that $\varepsilon < t$ implies $|y| >\frac{1}{2}|x|$ when $|x-y|<\varepsilon$. If there were, we could pick a sufficiently large $k$ so that $\varepsilon=p^{-k}$. Next I pick $x=p^{k+1}$, $y=p^{k+2}$ that satisfies $|x-y|=p^{-k-1} < \varepsilon$ and yet, $p^{-k-2}>\frac{1}{2}p^{-k-1}$ implies $2>p$ which is false.
Additionally as a suggestion to amend your question, for every fixed $x \in \mathbb{Q}^\times$ we can pick $\varepsilon<|x|$, and this will give us $|y|=|y-x+x|\le\max(|x-y|,|x|)$ by the ultrametric inequality. However because $|x-y|<\varepsilon < |x|$ we have that the ultrametric inequality's strong property upgrades it into an equality because $|x-y| \ne |x|$. So it becomes $|y|=|y-x+x|=\max(|x-y|,|x|)=|x|$. Now that we know $|y|=|x|$ we easily see $|y|>\frac{1}{2}|x|$.