= Solution
Let $\Psi\ne0$ and take the reduced <positive operator> $\rho_1=\operatorname{Tr}_2|\Psi\rangle\langle\Psi|$. Choose orthonormal <eigenvectors> $\xi_j$ for its positive <eigenvalues> $\lambda_j$. Define
$$
v_j=(\langle\xi_j|\otimes I)\Psi,\qquad c_j=\sqrt{\lambda_j},\qquad\eta_j=v_j/c_j.
$$
The defining identity of the <partial trace> gives $\langle v_i,v_j\rangle=\lambda_j\delta_{ij}$, so the $\eta_j$ are orthonormal. Expansion in a full eigenbasis of $\rho_1$ yields $\Psi=\sum_j\xi_j\otimes v_j$; zero-<eigenvalue> terms have zero squared norm and contribute nothing. Thus
$$
\boxed{\Psi=\sum_jc_j|\xi_j\rangle|\eta_j\rangle,\qquad c_j>0.}
$$
The zero vector has the empty decomposition. Unit <normalization> would give $\sum_jc_j^2=1$, but no <normalization> is needed for the theorem itself.
For the given vector, the <coefficient matrix> is $M=\bigl(\begin{smallmatrix}1&0\\0&1\\1&1\end{smallmatrix}\bigr)$. Its second reduced operator is $M^*M=\bigl(\begin{smallmatrix}2&1\\1&2\end{smallmatrix}\bigr)$, whose <eigenvalues> are $3,1$ and <eigenvectors> $\eta_+=(E_1+E_2)/\sqrt2$, $\eta_-=(E_1-E_2)/\sqrt2$. The corresponding normalized first-factor vectors are
$$
\xi_+=\frac{e_1+e_2+2e_3}{\sqrt6},\qquad\xi_-=\frac{e_1-e_2}{\sqrt2}.
$$
Their norms are one and their <inner product> is zero. Hence
$$
\boxed{\psi=\sqrt3\,|\xi_+\rangle|\eta_+\rangle+|\xi_-\rangle|\eta_-\rangle.}
$$
Expanding the two terms returns the original three contributions. The vector has squared norm $4$; if a normalized state is wanted, divide by two and the <Schmidt coefficients> become $\sqrt3/2$ and $1/2$.
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