Let $p \in \mathbb{R}^2$ and $v \in \mathbb{R}^2\backslash\{0\}$. Show that $$B(p+sv,s|v|) \subset B(p+tv,t|v|), \quad 0 < s < t$$ and that $$\bigcup_{t>0}B(p+tv,t|v|) = \{q \in \mathbb{R}^2 \mid (q-p)\cdot v> 0 \}.$$
Here, $B(c,r)$ is the open disk of radius $r$ centered at $c$.
I feel like I have to point out that normally, I would show some form of attempt. However, for this particular question, I was unable to find a starting point to solve this question. I was wondering if such problems should be approached purely algebraically or perhaps we should look at it geometrically? Any tips or hints?
My initial idea was something along this line:
Suppose we have $q \in B(p+sv,s|v|)$, that is, $|q - (p + sv)| \leq s|v|$, show that $|q -(p + tv)|<t|v|$ but I am unable to go anywhere with this.
The inclusion is straightforward: if $x \in B(p+sv,s|v|)$ then $|x - p - sv| < s|v|$. The triangle inequality gives you $$|x- p - tv| = |x-p-sv - (t-s)v| \le |x-p-sv| + |(t-s)v| < s|v| + (t-s)|v| = t|v|$$ because $t-s > 0$. Thus $|x-p-tv| < t|v|$ so that $x \in B(p+tv,t|v|)$, implying $$ B(p+sv,s|v|) \subset B(p+tv,t|v|).$$
Now suppose that $q \in B(p+tv,t|v|)$ for some $t > 0$. This means that $$|q-p-tv| < t|v|$$ so upon squaring and expanding you get $$|q-p|^2 - 2t (q-p) \cdot v + t^2 |v|^2 < t^2 |v|^2.$$ This reduces to $$ \tag{1}(q-p) \cdot v > \frac{|q-p|^2}{2t}$$ so that the inner product is positive.
Conversely, if you have a point $q$ satisfying $(q-p)\cdot v > 0$, choose $t > 0$ large enough so that (1) holds. You can reverse the steps above to find that $q \in B(p+tv,t|v|)$ so that $q$ belongs to the union as well.