 Solution

Construct the vectors inductively by the Gram-Schmidt process, proving its needed properties directly. Suppose $w_1,\dots ,w_{j-1}$ form an orthonormal basis of $U_{j-1}$, and putFor every $i<j$, taking an inner product with $w_i$ gives $⟨r_j,w_i⟩=0$. Also $r_j\ne 0$, because otherwise $u_j$ would lie in the linear span of its predecessors, contrary to linear independence. DefineIt lies in $U_j$, has unit norm, and is orthogonal to every $u_i$ with $i<j$, since those vectors lie in $U_{j-1}$. Writing $u_j=r_j+{\sum }_{k<j}⟨u_j,w_k⟩w_k$ gives $⟨w_j,u_j⟩=\Vert r_j\Vert >0$. The same identity shows that $w_1,\dots ,w_j$ span $U_j$, completing the induction, including the empty sum at $j=1$.

For uniqueness, any candidate $w\in U_j$ expands in the basis $w_1,\dots ,w_{j-1},r_j$. Orthogonality to every earlier $u_i$, hence to every earlier $w_i$, makes its first $j-1$ coefficients zero. Thus $w=t{r}_j$. Unit norm forces $t=\pm 1/\Vert r_j\Vert$, and $⟨w,u_j⟩>0$ forces the positive sign. This proves uniqueness in the Gram-Schmidt process with positive normalization without assuming a projection theorem or any other major orthogonalization result.