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A module retraction onto a submodule $U\stackrel{i}{\to }X$ is a module homomorphism $r:X\to U$ with $ri=I_U$. It gives $X=i(U)\oplus \ker r$: write $x=i(r(x))+(x-i(r(x)))$. Thus a nonzero proper submodule admitting a retraction contradicts indecomposability. Extending a projection from a larger submodule can produce such a retraction through the long exact sequence of Ext groups.

The standard projective resolution of a quiver representation has length one, even for quivers with oriented cycles. Thus every path algebra is a hereditary ring, and higher extension groups vanish. Applying a long exact sequence of Ext groups to $0\to I\to K\to C\to 0$ gives a surjection ${\mathrm{Ext}}^1(K,W)\twoheadrightarrow {\mathrm{Ext}}^1(I,W)$, since the next term ${\mathrm{Ext}}^2(C,W)$ is zero. This is the hereditary step in the Ringel lemma on bricks.

The Fitting lemma supplies a nonzero nilpotent endomorphism $f$ of a non-brick indecomposable $X$. Minimize its nonzero rank; nilpotence and minimality give $f^2=0$. Set $I=\mathrm{im}f\subset K=\ker f={⨁}_j{K}_j$. Choose a nonzero component $u:I\to K_j$. The square-zero endomorphism $X\stackrel{f}{\to }I\stackrel{u}{\to }{K}_j\hookrightarrow X$ has rank at least $\dim I$, so $u$ is injective.

If ${\mathrm{Ext}}^1(I,K_j)=0$, the projection $K\to K_j$ extends to $X\to K_j$ by the long exact sequence of Ext groups, giving a module retraction and contradicting indecomposability. Thus this extension group is nonzero. Since path algebras are hereditary, $u$ induces a surjection ${\mathrm{Ext}}^1(K_j,K_j)\twoheadrightarrow {\mathrm{Ext}}^1(I,K_j)$. Hence $K_j$ is a proper indecomposable submodule with self-extensions. Iterate until a brick module is reached; dimensions strictly decrease.

This minimal-rank argument is given in section 2 of William Crawley-Boevey's quiver lectures.