I am reading Brian Hall's Lie Groups, Lie Algebras, and Representation Theory. His Prop. 4.36 reads
If $G$ is a compact matrix Lie group, $G$ has the complete reducibility property.
The "complete reducibility property" means that every finite representation (of $G$) is completely reducible (i.e., is equivalent to a direct sum of irreducible representations). His proof invokes Haar measure, which he (admittedly) does not prove the existence of. He also uses (again without proof) that the Haar measure of a compact group is finite.
Is there a proof that does not require Haar measure or integration on manifolds?
In particular, the fact that we are only interested in compact matrix Lie groups should simplify things. I found some sources that suggest that existence of Haar measure is "trivial" to establish on Lie groups, but this requires familiarity with volume forms on manifolds, and I would prefer to even avoid this.
Question: "Is there a proof that does not require Haar measure or integration on manifolds?"
Answer: If your group $G$ is algebraic and semi-simple there is an elementary and algebraic proof in "Fulton/Harris - Representation theory a first course", Proposition C.15 using semi simple Lie algebras.
Hence if your Lie group is algebraic there is for every finite dimensional $G$-module $W$ a decomposition
$$W \cong W_1 \oplus \cdots \oplus W_k$$
where $W_i$ are irreducible $G$-modules.
When the book speaks about a "complementary $G$-module" $W'$ this means $V \cong W \oplus W'$. Hence by induction you get the above result.
https://en.wikipedia.org/wiki/Simple_Lie_group#Compact
Note: The book of FU speaks of complex representations. It may be the proof in FU can be adapted to your situation: That every finite dimensional real representation $W$ of $G$ decompose into a direct sum of irreducible representations.