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Here's a pretty easy direct argument. Let $X$ be a unitary representation of a compact group $G$. We note that for any finite-rank operator $T$ on $X$, $g\mapsto gTg^{-1}$ is a norm-continuous map $G\to B(X)$. This is because $T$ is a sum of operators of the form $\langle -, u\rangle v$, which conjugate to $\langle -, gu\rangle gv$, and $g\mapsto gv$ is norm-continuous for any fixed $v$ (the map $G\to U(X)$ is strong operator continuous).

Now let $T$ be any finite-rank positive operator on $X$. By averaging the conjugates of $T$ over $G$, we obtain an invariant positive operator $\tilde{T}$. By the continuity noted above and compactness of $G$, $\tilde{T}$ can be approximated in norm by finite "Riemann sums" of conjugates of $T$, and is thus compact.

If $X$ is irreducible, $\tilde{T}$ has to be a multiple of the identity. Since $\tilde{T}$ is compact, it follows that $X$ is finite-dimensional. More generally, eigenspaces of $\tilde{T}$ give finite-dimensional subrepresentations of any representation, and it follows easily that every unitary representation is a sum of irreducible representations (which are finite-dimensional).
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Here's a pretty easy direct argument. Let $X$ be a unitary representation of a compact group $G$. We note that for any finite-rank operator $T$ on $X$, $g\mapsto gTg^{-1}$ is a norm-continuous map $G\to B(X)$. This is because $T$ is a sum of operators of the form $\langle -, u\rangle v$, which conjugate to $\langle -, gu\rangle gv$, and $g\mapsto gv$ is norm-continuous for any fixed $v$ (the map $G\to U(X)$ is strong operator continuous).

Now let $T$ be any finite-rank projection positive operator on $X$. By averaging the conjugates of $T$ over $G$, we obtain an invariant projection positive operator $\tilde{T}$. By the continuity noted above and compactness of $G$, $\tilde{T}$ can be approximated in norm by finite "Riemann sums" of conjugates of $T$, and is thus a compactoperator.Any compact projection

If $X$ is finite-rank.

Thus any representation irreducible, $\tilde{T}$ has to be a finite-dimensional subrepresentationmultiple of the identity. In particularSince $\tilde{T}$ is compact, an irreducible representation must be it follows that $X$ is finite-dimensional.
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Here's a pretty easy direct argument. Let $X$ be a unitary representation of a compact group $G$. We note that for any finite-rank operator $T$ on $X$, $g\mapsto gTg^{-1}$ is a norm-continuous map $G\to B(X)$. This is because $T$ is a sum of operators of the form $\langle -, u\rangle v$, which conjugate to $\langle -, gu\rangle gv$, and $g\mapsto gv$ is norm-continuous for any fixed $v$ (the map $G\to U(X)$ is strong operator continuous).

Now let $T$ be any finite-rank projection on $X$. By averaging the conjugates of $T$ over $G$, we obtain an invariant projection $\tilde{T}$. By the continuity noted above and compactness of $G$, $\tilde{T}$ can be approximated in norm by finite "Riemann sums" of conjugates of $T$, and is thus a compact operator. Any compact projection is finite-rank.

Thus any representation has a finite-dimensional subrepresentation. In particular, an irreducible representation must be finite-dimensional.