(Addressing only the title question.) There is a short proof avoiding Peter-Weyl and the theory of compact operators. It is due to Nachbin and is reproduced in Hewitt-Ross, Abstract Harmonic Analysis 1, p. 344. A slight simplification of it runs as follows: Pick a unit vector $v$ in your representation space $V$. Schur's lemma gives $$\int_G gv(gv,\cdot)dg = \lambda1 \tag 1$$ where $\lambda = \int_G|(v,gv)|^2dg>0$ (sandwiching (1) with $(v,\cdot v)$). Now let $W\subset V$ be finite-dimensional, and write $E=E^2$ for the orthogonal projection $V\to W$. We get $$\int_GEgv(gv,E\cdot)dg = \lambda E, \tag 2$$ whence, taking traces in (2), $\lambda\dim(W)=\int_G||Egv||^2dg\leqslant\operatorname{vol}(G)$. Thus, the dimension of any finite-dimensional subspace is bounded, as was to be shown.
(Addressing only the title question.) There is a short proof avoiding Peter-Weyl and the theory of compact operators. It is due to Nachbin and reproduced in Hewitt-Ross, Abstract Harmonic Analysis 1, p. 344. A slight simplification of it runs as follows: Pick a unit vector $v$ in your representation space $V$. Schur's lemma gives $$\int_G gv(gv,\cdot)dg = \lambda1 \tag 1$$ where $\lambda = \int_G|(v,gv)|^2dg>0$ (sandwiching (1) with $(v,\cdot v)$). Now let $W\subset V$ be finite-dimensional, and write $E=E^2$ for the orthogonal projection $V\to W$. We get $$\int_GEgv(gv,E\cdot)dg = \lambda E, \tag 2$$ whence, taking traces in (2), $\lambda\dim(W)=\int_G||Egv||^2dg\leqslant\operatorname{vol}(G)$. Thus, the dimension of any finite-dimensional subspace is bounded, as was to be shown.