We'll induct on the number of distinct prime factors in the group's order. As you've said, the base case is pretty easy. Now, suppose the statement has been proven for all abelian groups, $H$, such that $|H|$ has $n$ distinct prime factors.
Let $G$, then, be an abelian group with $n+1$ distinct prime factors. This means that we can factor $|G|$ as $p^kj$ where $p^k$ and $j$ are coprime, and $j$ has $n$ distinct prime factors. Let $P$ be the set of all elements in $G$ whose order divides $p^k$ and let $J$ be the set of all elements in $G$ whose order divides $j$.
It is easy to see that both $P$ and $J$ are subgroups of $G$. It is also easy to see, by Cauchy's theorem, that $|P|$ and $|J|$ must be coprime. This will be important later.
First, each element in $G$ can be expressed as a sum of an element in $P$ and element in $J$. This is relatively easy to show. For any element, $x\in G$, we have $|G|\cdot x=0$, i.e. $(p^kj)\cdot x=0$. But this can be rewritten in a couple of ways- we can write it as $j\cdot (p^k\cdot x)=0$ which gives us that $\operatorname{ord}(p^k\cdot x)|j$.
Similarly, $\operatorname{ord}(j\cdot x)|p^k$. So, by the definitions of $P$ and $J$, for any $x\in G$, $p^k\cdot x\in J$ and $j\cdot x\in P$.
But, since $p^k$ and $j$ are coprime we can pick $a$ and $b$ so that $ap^k+bj=1$, which means $a\cdot(p^k\cdot x)+b\cdot(j\cdot x)=(ap^k)\cdot x+(bj)\cdot x=(ap^k+bj)\cdot x=1\cdot x=x$.
By what we have already shown, $a\cdot(p^k\cdot x)\in J$ and $b\cdot(j\cdot x)\in P$, so each $x\in G$ can be written as the combination of an element in $P$ and an element in $J$.
Also, since the order of any element in $P$ is coprime to the order of any non-zero element in $J$ (by the definitions of $P$ and $J$), we have $P\cap J=\{0\}$.
This finally gives us $G=P\oplus J$. Given that: $|P|$ and $|J|$ are coprime, $|P|=p^s$ for some $s$ (by Cauchy's theorem), and that $|G|=|P||J|$ (because $G$ is the direct sum of $P$ and $J$), we have $|P|=p^k$ and $|J|=j$. Now, we can apply the induction hypothesis on $J$ to expand it as a direct sum of groups of prime order in distinct primes, and, with $G=P\oplus J$, we are done.
For your example, $T$ is bounded, so to apply 4.3.8, it suffices to show $T$ is pure.
Let $t\in T$, and suppose $n{\,\mid\,}t$ in $G$.
We want to show $n{\,\mid\,}t$ in $T$.
Since $n{\,\mid\,}t$ in $G$, we have $t=ng$ for some $g\in G$.
Since $t\in T$, we have $mt=0$ for some positive integer $m$, so
$$(mn)g=m(ng)=mt=0$$
hence $g\in T$
Thus $n{\,\mid\,}t$ in $T$, so $T$ is pure.
Now consider the general case . . .
Thus suppose $T=D\,{\large{\oplus}}B$, where $D$ is divisible and $B$ is bounded.
Our goal is to show that $T$ is a direct summand of $G$.
Since $D$ is divisible, it follows by 4.1.3 that $G=D\,{\large{\oplus}}H$ for some $H\le G$.
Let $S$ be the torsion subgroup of $H$.
Since $B$ is bounded, we have $mB=0$ for some positive integer $m$.
Let $s\in S$.
Clearly $s\in T$ so $s=d+b$, where $d\in D$ and $b\in B$.$\;$Then
\begin{align*}
&
s=d+b\\[4pt]
\implies\;&
ms=md+mb\\[4pt]
\implies\;&
ms=md\\[4pt]
\implies\;&
ms\in D\\[4pt]
\implies\;&
ms=0\;\;\;\text{[since $ms\in S\subseteq H$ and $D\cap H=0$]}\\[4pt]
\end{align*}
Thus $mS=0$ so $S$ is bounded.
Since $S$ is the torsion subgroup of $H$, it follows (as in the first part of this post) that $S$ is pure in $H$, hence since $S$ is bounded, it follows by 4.3.8 that $S$ is a direct summand of $H$.
Thus we have $H=S\,{\large{\oplus}}K$ for some $K\le H$, hence we can rewrite $G=D\,{\large{\oplus}}H$ as $G=D\,{\large{\oplus}}S\,{\large{\oplus}}K$.
Claim $T=D\,{\large{\oplus}}S$.
The inclusion $D\,{\large{\oplus}}S\subseteq T$ is clear.
For the reverse inclusion, let $t\in T$.
Since $G=D\,{\large{\oplus}}S\,{\large{\oplus}}K$, we have $t=d+s+k$, where $d\in D$, $s\in S$, and $k\in K$.
Then from $k=t-d-s$, it follows that $k$ has finite order, hence $k=0$.
Thus $t=d+s\in D\,{\large{\oplus}}S$, which yields $T\subseteq D\,{\large{\oplus}}S$.
Hence $T=D\,{\large{\oplus}}S$, as claimed.
Then we can rewrite $G=D\,{\large{\oplus}}S\,{\large{\oplus}}K$ as $S=T\,{\large{\oplus}}K$, so $T$ is a direct summand of $G$ as was to be shown.
Best Answer
If you want a direct proof, the proof of the classification theorem is not so hard anyways.
Indeed, clearly the two subgroups in question have trivial intersection, and they're both normal since $\mu$ is abelian. Moreover if $x\in \mu$ has order $n=p^\alpha m$ with $m\land p = 1$, then write $um + vp^\alpha = 1$ with Bezout's theorem, so that $x= x^{um+vp^\alpha} = x^{um}x^{vp^\alpha}$ and you can easily check that $x^{um}$ has order $p^\alpha$ and $x^{vp^\alpha}$ has order $m$, that is, order prime with $p$.
The decomposition follows.