Every non-zero submodule of $R_R$ has an indecomposable direct summand: True when $R$ is von Neumann regular? Let's say that a (right) module $M$ is well complemented if every non-zero submodule of $M$ has an indecomposable direct summand (by the way, is there a better or more standard name for this property?). For instance, every module of finite uniform dimension is well complemented.

Question. Is the regular right module $R_R$ of a von Neumann regular ring $R$ well complemented?

As a recall, a ring $R$ is von Neumann regular if, for every $x \in R$, there exists $y \in R$ such that $x = xyx$.
 A: The answer is no.  Take a compact totally disconnected space $X$ with no isolated points, like the Cantor set.  Let $K$ be any field and let $R$ be the ring of locally constant functions $f\colon X\to K$ with pointwise operations.  This is a commutative von Neumann regular ring.  The idempotents of $R$ are precisely the characteristic functions $1_K$ of clopen sets $K$.  An orthogonal decomposition of $1_K$ into idempotents corresponds to writing $K$ as a disjoint union of clopen sets.  Since $X$ has no isolated points, if $K$ is a nonempty clopen set, there are $x\neq y\in K$.  Then we can find a clopen subset $K'$ of $K$ with $x\in K'$ and $y\notin K'$.  Thus $1_K = 1_{K'}+1_{K\setminus K'}$ is a decomposition into orthogonal idempotents.  If follows that $R$ has no primitive idempotents and hence no indecomposable summands (as an indecomposable summand is of the form $eR$ with $e$ primitive).
A: No, a free Boolean algebra $R$ on an infinite cardinal $\kappa$ (e.g., if $\kappa = \aleph_0$, $R$ is the Cantor algebra), is a commutative von Neumann regular ring which is not well complemented as an $R$-module.
