The idea is roughly as follows: Let $G$ be a compact Lie group and $G_0$ be the connecting component of the identity element. Then $BG_0 \to BG$ is a covering and $$H(BG;\mathbb{R}) H^\ast(BG;\mathbb{R}) = H^*(BG_0;\mathbb{R})^{\Gamma}$$ where $\Gamma = G/G_0$ is a finite group. Let $T$ be a maximal torus of $G_0$. Then one shows that $$H(BG_0;\mathbb{R}) H^\ast(BG_0;\mathbb{R}) = H^*(BT;\mathbb{R})^W$$ H^\ast(BT;\mathbb{R})^W$$where W is the Weyl group of G_0. Thus H(BG;\mathbb{R}) H^\ast(BG;\mathbb{R}) can be identified with a subring of H^{*}(BT;\mathbb{R}) H^\ast(BT;\mathbb{R}) that is a polynomial ring with generators of degree two. So H^{odd}(BG;\mathbb{R}) = 0 follows. Editorial remark: The three plain H's should be H^*. No idea why the asterisk doesn't work at those places. 1 For a reference see "Hsiang: Cohomology theory of topological transformation groups" (chapter III, §1). The results of the book that are relevant for your question can also be found in the following paper: http://www.math.uwo.ca/~rgonzal3/qfy.pdf (cf. Remark 9, Lemma 5). The idea is roughly as follows: Let G be a compact Lie group and G_0 be the connecting component of the identity element. Then BG_0 \to BG is a covering and$$H(BG;\mathbb{R}) = H^*(BG_0;\mathbb{R})^{\Gamma}$$where \Gamma = G/G_0 is a finite group. Let T be a maximal torus of G_0. Then one shows that$$H(BG_0;\mathbb{R}) = H^*(BT;\mathbb{R})^W where $W$ is the Weyl group of $G_0$. Thus $H(BG;\mathbb{R})$ can be identified with a subring of $H^{*}(BT;\mathbb{R})$ that is a polynomial ring with generators of degree two. So $H^{odd}(BG;\mathbb{R}) = 0$ follows.
Editorial remark: The three plain $H$'s should be $H^*$. No idea why the asterisk doesn't work at those places.