$\mathfrak{so}(3)$ with the bracket is isomorphic to $(\mathbb{R}^3, \times)$

Step 1 (basis). Every skew-symmetric $X \in \mathfrak{so}(3)$ has the form $X_u = \begin{pmatrix} 0 & -u_3 & u_2 \\ u_3 & 0 & -u_1 \\ -u_2 & u_1 & 0 \end{pmatrix}$ for a unique $u=(u_1,u_2,u_3) \in \mathbb{R}^3$ (three free entries above the diagonal determine the rest), and $u \mapsto X_u$ is manifestly linear and bijective, so it is a linear isomorphism of $3$-dimensional vector spaces.

Step 2 (action on vectors). A direct computation shows $X_u v = u \times v$ for every $v \in \mathbb{R}^3$, i.e. $X_u$ acts on $\mathbb{R}^3$ exactly as "cross with $u$".

Step 3 (bracket matches cross product). Using Step 2 twice, for any $v$: $[X_u,X_v]w = X_u(X_v w) - X_v(X_u w) = u\times(v\times w) - v\times(u\times w)$. Applying the vector triple-product identity $a\times(b\times c) = b(a\cdot c) - c(a\cdot b)$ to both terms and simplifying, the right side collapses to $(u\times v)\times w = X_{u\times v} w$ for every $w$, hence $[X_u,X_v] = X_{u\times v}$ as operators — exactly $\mathfrak{so}(3) \cong (\mathbb{R}^3, \times), \qquad [X_u, X_v] = X_{u \times v}$.

Step 4 (conclusion). Since $u \mapsto X_u$ is a linear bijection that turns the cross product into the matrix bracket, it is a Lie algebra isomorphism $(\mathbb{R}^3,\times) \cong \mathfrak{so}(3)$, as claimed.