How to expand a wavefunction of a two-particle system in one dimension in some basis without operations associated with the tensor product?

Question

In Shankar's Principle of Quantum Mechanics, Section 10.1, part The Direct Product Revisited (he calls tensor products direct products), he attempts to show that a two-particle state space is the tensor product of two one-particle state spaces. He begins by letting $\Omega_1^{(1)}$ be an operator on the state space $\Bbb V_1$ of a particle in one dimension, whose nondegenerate eigenfunctions $\psi_{\omega_1}(x_1)$ form a complete basis, and similarly, letting $\psi_{\omega_2}(x_2)$ form a basis for the state space of a second particle. He then states that if a function $\psi(x_1,x_2)$ that represents an abstract vector $|\psi\rangle$ from the state space $\Bbb V_{1\otimes2}$ of a system consisting of both particles has $x_1$ fixed at some value $\bar x_1$, then $\psi$ becomes a function of $x_2$ alone and may be expanded as $$\psi(\bar x_1,x_2)=\sum_{\omega_2}C_{\omega_2}(\bar x_1)\psi_{\omega_2}(x_2)\tag{1}$$ where $$C_{\omega_2}(\bar x_1)=\sum_{\omega_1}C_{\omega_1\omega_2}\psi_{\omega_1}(\bar x_1)\tag{2}$$ Substituting Equation $(2)$ into Equation $(1)$ and dropping the bar on $\bar x_1$, he states that the resulting expansion $$\psi(x_1,x_2)=\sum_{\omega_1}\sum_{\omega_2}C_{\omega_1\omega_2}\psi_{\omega_1}(x_1)\psi_{\omega_2}(x_2)\tag{3}$$ imply that $\Bbb V_{1\otimes2}=\Bbb V_1\otimes\Bbb V_2$ for $\psi_{\omega_1}(x_1)\times\psi_{\omega_2}(x_2)$ is the same as the inner product between $|x_1\rangle\otimes|x_2\rangle$ ($|x_1\rangle$ and $|x_2\rangle$ are the position basis vectors of $\Bbb V_1$ and $\Bbb V_2$ respectively) and $|\omega_1\rangle\otimes|\omega_2\rangle$ ($|\omega_1\rangle$ and $|\omega_2\rangle$ are the basis eigenvectors of $\Omega_1$ on $\Bbb V_1$ and $\Omega_2$ on $\Bbb V_2$ respectively).

Question: How does Equation $(1)$ follow from fixing $x_1$ in $\psi(x_1,x_2)$ as $\bar x_1$? Using the simultaneous eigenbasis $|\omega_1\omega_2\rangle$ of the operators $\Omega_1$ and $\Omega_2$ on $\Bbb V_{1\otimes2}$, $$\psi(\bar x_1,x_2)=\langle\bar x_1x_2|\psi\rangle=\sum_{\omega_1}\sum_{\omega_2}\langle\omega_1\omega_2|\psi\rangle\langle\bar x_1x_2|\omega_1\omega_2\rangle=\sum_{\omega_1}\sum_{\omega_2}\langle\omega_1\omega_2|\psi\rangle\psi_{\omega_1\omega_2}(\bar x_1,x_2).\tag{4}$$If the author intends $C_{\omega_1\omega_2}$ to mean $\langle\omega_1\omega_2|\psi\rangle$, what is the reason (besides the tensor product since he is trying to show that $\Bbb V_{1\otimes2}=\Bbb V_1\otimes\Bbb V_2$) for $\psi_{\omega_1\omega_2}(\bar x_1,x_2)=\psi_{\omega_1}(\bar x_1)\psi_{\omega_2}(x_2)$?