not understanding the proof of jacobson-semisimple and DCC on principals implying semisimpleness Let $R$ be a ring(not necessarily commutative) s.t. $R$ is jacobson-semisimple(i.e. Jacobson radical is $0$.) and satisfying DCC on principal left ideals. The claim is to deduce $R$ is semi-simple. I am treating ideals as left ideals. 
I am basically quoting Lam's Introduction to Noncommutative Rings proof but I do not see the last part of the proof. 
(1) Every ideal contains a minimal ideal. This follows from DCC properties on principal left ideals and picking any element of the ideal to form a principal ideal. 
(2) Every minimal ideal is a direct summand. Say $(a)$ is a minimal ideal. This has to be simple as left module as well by minimality. Since $R$ is jacobson semisimple($Jac(R)=0$), we must have $\cap m=0$ where $m$ runs through all maximal left ideals.(Note that intersection of left and right does not make difference under symmetrization procedure. Left intersection is sufficient.) So $(a)\cap (\cap m)=0$. In particular, we can find $m$ s.t. $(a)+m=1$ by $m$ maximal and $(a)\cap(\cap m)=0$. Then $(a)\cap(\cap m)=0\implies (a)\oplus m=R$ as left $R$ module. 
Assume $R$ is not semi-simple. There is an ideal $I$ which is not a direct summand of $R$.(I am not certain about this reasoning.) From (2) and (1), pick a minimal ideal $(a_1)$ of $I$, one has $R=(a_1)\oplus B$ with some ideal $B$ but we know $B$ is maximal ideal $m$ by (2)'s direction construction. Apply $(1)$ again to $m$ to obtain $(a_2)\subset m$ and apply $(2)$ to see $m=(a_2)\oplus m_1$. Then one will obtain a descending chain $m\supset m_1\supset m_2\supset\dots$. 
Then the book claims that $m,m_1,m_2,\dots$ are direct summands of $R$ as left module.(This is clear.) So $m,m_1,m_2,\dots$ are principal left ideals.
$\textbf{Q:}$ Since the book wants to prove the statement by contradiction, he assumes $R$ is not semi-simple. The book really starts with picking a minimal left ideal $B$ and getting $B\oplus A=R$. Then the argument flows as iteration against $A$. Do I even need to refer to some ideal that is not a direct summand of $R$ here?
$\textbf{Q':}$ Why $m,m_1,m_2,\dots$ are principal left ideals?(I have $R$ decomposed into sums of principals that are simple.
I think the proof is faily convoluted to me. In my view, I could have started by considering $R=\oplus (a_i)$ with $(2)$ and $(1)$ construction for iteration and using for simple $(a),(b)$ ideals, $(a)\cap (b)=0$ or $(a)\cap (b)=(a)=(b)$. Then this will decompose $R=\oplus_i (a_i)$. with $(a_i)$ simple. Now 1 is finite liner combination of $a_i$. Hence, there is only finite list of index $i$ left. So $R$ is a finite direct summand of simple modules. Hence $R$ is semisimple. 
Ref. Lam, A first course in noncommutative rings, Chpt 2, Sec 4. (4.14) Thm proof.
 A: Second Question
I'll begin with Q', since it is easier to answer.
If $I$ is a direct summand of $R$, then it has a surjective projection map $\pi : R\to I$, and is thus cyclic (and therefore principal).
First Question
I think it's just a bit confusingly worded.
The proof I see finishes like this, and I think that this is what the proof you've quoted is saying.
Proof.
We should iteratively end up with modules $A_n\oplus m_{n-1}=R$, with $A_n\subseteq I$, by always choosing $a_n\in I$. We can do that for the following reason.
If $I\cap m_j=0$, then since $A_{j+1}\subseteq I$, and $A_{j+1}+m_j=R$, then we have $I\oplus m_j=R$, contradicting the choice of $R$. Thus $I\cap m_j\ne 0$, so it contains a minimal left ideal $(a_{j+2})$. (Side note, what's with the choice of indices here).
Since $I\cap m_j$ is never $0$, we can always extend the chain, so the $m_j$ form an infinite descending sequence of ideals that doesn't stabilize.
However, since the $m_n$ are direct summands of $R$, they are principal, and thus stabilize.
Contradiction.
