[First proof] I will address only the first part. Suppose $G$ is a compact group and $\pi : G \rightarrow U(H)$ a unitary irreducible representation on a Hilbert space $H$. Suppose $\phi$ a continuous function on $G$. Then it is easy to show that $\pi (\phi)$ is a compact operator on $H$.
Since the OP has asked for clarification let me state: convolution by continuous functions on $L^2(G)$ where $G$ a COMPACT group, is a compact operator. This fact is used in all proofs of Peter-Weyl. The proof is by showing that the convolution is an $L^2$ kernel, and any $L^2$ function on $G\times G$ may be approximated in $L^2$ by simple functions (linear combination of char functions of the form $E\times F$ where $E,F$ ae measurable subsets of $G$), each of these simple kernels have finite dimensional image and are hence compact operators. This is standard material in any functional analysis book (in fact Kirillov's book).
What I have used is that for a compact group, any irreducible unitary representation is a sub of the regular representation (see the comments), and hence $\pi (\phi)$ ,which is a convolution by $\phi$ is a compact operator.
If $(\phi _{\epsilon} )$ is an approximate identity on $G$ consisting of continuous conjugation invariant real valued continuous functions on $G$, then $\pi (\phi _{\epsilon})$ is a sequence of compact operators (which are scalars because of irreducibility) converging weakly to the identity, and hence the identity operator is also compact. Therefore, $H$ is finite dimensional.
I mention this because this does not use the Peter-Weyl theorem (but uses ideas of the proof).
[Second Proof] If you use the Peter-Weyl theorem, then the proof of finite dimensionality is easy. Peter weyl says that $L^2(G)$ is a Hilbert space direct sum of irreducible FINITE dimensional representations of $G$. The algebraic direct sum $X$ is then a dense subspace of $L^2(G)$. If an infinite dimensional irrep $H$ existed, it would still be in $L^2(G)$ but by Schur orthogonality (another use of Schur's lemma) functions in $H$ would be orthogonal to functions in $X$, a contradiction, since $X$ is dense.
[Remark] if $G$ is already a closed subgroup of $U(n)$, and $V$ is the standard rep of $U(n)$ then by the Stone-Weierstrass theorem, $V$, $V^*$ and irreducible subs of their tensor powers, will all be in $L^2(G)$ and will be dense in the space of continuous functions on $G$. And hence these representation functions $Y$ will be dense in $L^2(G)$. So vectors in $H$ cannot be orthogonal to the elements of $Y$. All this is standard material in any book on the Peter-Weyl theorem (actually worked out in Kirillov's book) and I suggest that the OP looks at them
[Remark] The proof given by Francois Ziegler is the simplest, in my opinion. .
