$SO(3)$ is a subgroup of the general linear group, with $R^{-1}=R^{\mathsf{T}}$

Since $(I_3{)}^{\mathsf{T}}{I}_3=I_3{I}_3=I_3$ and $\det I_3=1$ by [L1] and [L3], the identity matrix lies in $SO(3)$.

For the product, $(RS{)}^{\mathsf{T}}(RS)=S^{\mathsf{T}}{R}^{\mathsf{T}}RS=S^{\mathsf{T}}(R^{\mathsf{T}}R)S=S^{\mathsf{T}}{I}_{3}S=S^{\mathsf{T}}S=I_3$ by [L1], [L2] and the defining relations for $R,S$; and $\det (RS)=\det R\det S=1$ by [L3]. Hence $RS\in SO(3)$.

For $R\in SO(3)$ the relation $R^{\mathsf{T}}R=I_3$ exhibits $R^{\mathsf{T}}$ as a left inverse of $R$. Since $\det R=1\mathrm{\ne }0$, the matrix $R$ is invertible by [L4]; multiplying the relation $R^{\mathsf{T}}R=I_3$ on the right by the two-sided inverse $R^{-1}$ and using [L1] gives $R^{\mathsf{T}}=R^{-1}$; hence also $R{R}^{\mathsf{T}}=R{R}^{-1}=I_3$. Moreover $R^{\mathsf{T}}\in SO(3)$: $(R^{\mathsf{T}}{)}^{\mathsf{T}}{R}^{\mathsf{T}}=R{R}^{\mathsf{T}}=I_3$ by [L2], and $\det R^{\mathsf{T}}=1$ because $\det R^{\mathsf{T}}\det R=\det (R^{\mathsf{T}}R)=\det I_3=1$ by [L2], [L3] and $\det R=1$.

By steps 1.1, 1.2 and 2.1 the set $SO(3)$ contains $I_3$ and is closed under multiplication and inversion; since it is a subset of $G{L}_3(\mathbb{R})$ by step 2.1 and $G{L}_3(\mathbb{R})$ is a group under multiplication by [L4], $SO(3)$ is a subgroup of $G{L}_3(\mathbb{R})$. This proves clause 2.