$SO(3)$ is a compact three-dimensional embedded submanifold of $M_3(\mathbb{R})$, with tangent space the matrices $R\mathrm{\Omega }$ for skew $\mathrm{\Omega }$

Set ${\mathrm{Sym}}_3(\mathbb{R}):=\{S\in M_3(\mathbb{R}):S^{\mathsf{T}}=S\}$ and identify it with ${\mathbb{R}}^6$ by its six entries with $i\le j$. Define $f:M_3(\mathbb{R})\to {\mathrm{Sym}}_3(\mathbb{R})$ by $f(A):=A^{\mathsf{T}}A$. Its independent components are $f_{ij}(A)={\sum }_{k<3}{a}_{ki}{a}_{kj}$ for $i\le j$, finite sums of products of coordinate functions; each coordinate function is linear with constant first partials and zero higher partials, hence $C^k$ for every $k$. By [F2], $f$ is $C^k$ for every $k$, hence smooth. The target has dimension $6$ by the displayed coordinate identification.

For $A,H\in M_3(\mathbb{R})$ and real $t$, $f(A+tH)=(A^{\mathsf{T}}+t{H}^{\mathsf{T}})(A+tH)=A^{\mathsf{T}}A+t(A^{\mathsf{T}}H+H^{\mathsf{T}}A)+t^2{H}^{\mathsf{T}}H$, so the directional derivative in direction $H$ is $A^{\mathsf{T}}H+H^{\mathsf{T}}A$. Since $f$ is smooth, [F9] gives total differentiability and [F3] identifies $Df(A)H=A^{\mathsf{T}}H+H^{\mathsf{T}}A$. If $A\in O(3)$ then $A^{\mathsf{T}}A=I_3$ and the derivative is surjective onto ${\mathrm{Sym}}_3(\mathbb{R})$: for symmetric $S$ take $H:=AS\mathrm{/}2$, so that $A^{\mathsf{T}}H+H^{\mathsf{T}}A=A^{\mathsf{T}}AS\mathrm{/}2+S^{\mathsf{T}}{A}^{\mathsf{T}}A\mathrm{/}2=S\mathrm{/}2+S\mathrm{/}2=S$. The level is nonempty because $I_3\in O(3)$ by [F7]; hence $I_3$ is a regular value and [F1] gives that $O(3)=f^{-1}(I_3)$ is an embedded submanifold of $M_3(\mathbb{R})$ of dimension $9-6=3$ with $T_{A}O(3)=\ker Df(A)$ for $A\in O(3)$.

For $A\in O(3)$ the kernel is $\{AK:K^{\mathsf{T}}=-K\}$: if $K$ is skew then $Df(A)(AK)=A^{\mathsf{T}}AK+(AK{)}^{\mathsf{T}}A=K+K^{\mathsf{T}}=0$; conversely, if $A^{\mathsf{T}}H+H^{\mathsf{T}}A=0$, put $K:=A^{\mathsf{T}}H$, so that $K^{\mathsf{T}}=H^{\mathsf{T}}A=-A^{\mathsf{T}}H=-K$ and $H=AK$ because [F7] gives $A{A}^{\mathsf{T}}=I_3$. Combining with step 2.1, $T_{A}O(3)=\{AK:K^{\mathsf{T}}=-K\}$.

For $A\in O(3)$, [F7] gives $(\det A{)}^2=1$; hence $\det A>0$ if and only if $\det A=1$ by [F8]. Therefore $SO(3)=O(3)\cap {\det }^{-1}((0,\mathrm{\infty }))=O(3)\cap {\det }^{-1}(\{1\})$. By [F4] the determinant is continuous, $(0,\mathrm{\infty })$ is open and $\{1\}$ is closed, so both preimages are open respectively closed; hence $SO(3)$ is open and closed in $O(3)$. By [F5], $SO(3)$ has the restricted smooth structure of dimension $3$. Fix $R\in SO(3)$ and a slice chart of $O(3)$ at $R$ from step 2.1 and [F10]. Restrict its ambient domain to its intersection with the open set ${\det }^{-1}((0,\mathrm{\infty }))$. On this domain $SO(3)$ and $O(3)$ coincide, so the same chart is a slice chart for $SO(3)$, proving embeddedness in $M_3(\mathbb{R})$. In the restricted manifold charts the inclusion $j:SO(3)\hookrightarrow O(3)$ is the identity in three coordinates. The differential formula in [F10] sends each coordinate derivation to the corresponding coordinate derivation of $O(3)$; the coordinate-basis theorem makes $d{j}_R$ an isomorphism onto $T_{R}O(3)$. Composing with the ambient inclusion therefore gives exactly the same ambient tangent subspace as in step 3.1, namely $\{R\mathrm{\Omega }:{\mathrm{\Omega }}^{\mathsf{T}}=-\mathrm{\Omega }\}$. This proves clauses 1 and 2.

$SO(3)$ is closed in $M_3(\mathbb{R})$ as the intersection of the closed sets $O(3)=f^{-1}(\{I_3\})$ and ${\det }^{-1}(\{1\})$, using continuity of $f$ and of the determinant and closedness of singletons from [F4]. It is bounded: if $A\in SO(3)$ then the relation $A^{\mathsf{T}}A=I_3$ says that the columns of $A$ are unit vectors, so $\mathrm{\mid }{a}_{ij}\mathrm{\mid }\le 1$ for all $i,j$ and the Frobenius norm obeys $\mathrm{\parallel }A{\mathrm{\parallel }}_F^2={\sum }_{i,j}{a}_{ij}^2\le 9$. By [F6] a closed and bounded subset of ${\mathbb{R}}^9$ is compact, so $SO(3)$ is compact. This proves clause 3.