You don’t need to use the machinery in that PDF to prove it.
By definition the ball $B(x,r^-)$ is open. Suppose that $y\in X\setminus B(x,r^-)$. Then $d(x,y)\ge r$, and I claim that $B(y,r^-)$ is an open ball around $y$ disjoint from $B(x,r^-)$. Assuming the claim for a moment, it follows that $X\setminus B(x,r^-)$ is open and hence that $B(x,r^-)$ is closed.
To prove the claim, suppose that $z\in B(y,r^-)\cap B(x,r^-)$. Then $$d(x,y)\le\max\{d(x,z),d(z,y)\}<r\;,$$ since $d$ is an ultrametric, contradicting the choice of $y$.
This shows that every open ball is also closed. Now we’ll show that the closed ball $B(x,r)$ is also open. Let $y\in B(x,r)$ be arbitrary; I claim that $B(y,r^-)\subseteq B(x,r)$, from which it follows immediately that $B(x,r)$ is open.
To prove the claim, let $z\in B(y,r^-)$, so that $d(z,y)<r$. We also have, $d(x,y)\le r$ by the choice of $y$. Thus $$d(z,x)\le\max\{d(z,y),d(y,x)\}\le r\;,$$ so $z\in B(x,r)$.
Added: However, it’s important that you learn to work with partitions and equivalence relations: they’re pretty nearly ubiquitous in mathematics, and they’re very handy tools. For the basics see the Wikipedia articles on equivalence relations, partitions, and equivalence classes and/or this page.
If $\langle X,d\rangle$ is an ultrametric space, and $r$ is any positive real number, we define a relation $\sim_r$ on $X$ as follows: $$x\sim_r y\quad\text{ iff }\quad d(x,y)\le r\;.$$
Clearly $x\sim_r x$ for every $x\in X$, since $d(x,x)=0\le r$, so $\sim_r$ is reflexive. The ultrametric $d$ is a symmetric function, so $x\sim_r y$ iff $y\sim_r x$, and $\sim_r$ is therefore symmetric. Finally, if $x\sim_r y$ and $y\sim_r z$, then $$d(x,z)\le\max\{d(x,y),d(y,z)\}\le r\;,$$ so $x\sim_r z$, and $\sim_r$ is transitive. By definition, then $\sim_r$ is an equivalence relation.
Now $B(x,r)=\{y\in X:d(x,y)\le r\}=\{y\in x:x\sim_r y\}$; that last set is by definition the $\sim_r$-equivalence class of $x$, so for each $x\in X$ the $\sim_r$-equivalence class of $x$ is simply the closed ball $B(x,r)$. It is a general fact about equivalence relations that equivalence classes are either disjoint or identical. In this case that means that for any $x,y\in X$, either $B(x,r)\cap B(y,r)=\varnothing$, or $B(x,r)=B(y,r)$. To put it in slightly different language, the equivalence classes of any equivalence relation form a partition of the underlying set: they divide it into parts in such a way that every point is in exactly one part.
Similarly, we can define
$$x\sim_{r^-} y\quad\text{ iff }\quad d(x,y)<r\;$$
and show that it too is an equivalence relation on $X$, and that its equivalence classes are the open balls $B(x,r^-)$ for $x\in X$. This is what’s being used in that PDF: since the balls $B(x,r^-)$ for $x\in X$ divide up $X$ in such a way that each point is in exactly one of them, the complement of any one of them is the union of all the others and is therefore open, being a union of open sets. This means that each one of them must be closed.
Best Answer
Let $C\subseteq \Bbb R^n$ be convex, let $f\colon \Bbb R^n\to\Bbb R^n$ be an affine linear map (i.e., $f(x)=Ax+b$ with an $n\times n$ matrix $A$ and $b\in\Bbb R^n$). Then $f(C)$ is convex. Indeed, if $x,y\in f(C)$, say $x=f(u), y=f(v)$, then any convex combination $tx+(1-t)y$, $0\le t\le 1$, is also in $f(C)$, namely $$\begin{align}f(tu+(1-t)v)&=A(tu+(1-t)v)+b\\&=tAu+(1-t)Av+tb+(1-t)b\\&=t(Au+b)+(1-t)(Av+b)\\&=tx+(1-t)y.\end{align}$$
Now note that $B_r(x_0)$ is obtained from $B_1(0)$ by (linearly) scaling with $r$ and translating by $x_0$.