Given $C_c^{\infty}((a_i,b_i))$ for $i \in \{1,2\}.$ Is it then true that
$$C_c^{\infty}(a_1,b_1) \otimes C_c^{\infty} (a_2,b_2)$$ is dense in $H_0^1((a_1,b_1) \times (a_2,b_2))$? Clearly, by definition $C_c^{\infty}((a_1,b_1) \times (a_2,b_2))$ is dense in $H_0^1((a_1,b_1) \times (a_2,b_2))$. But is this also true for the tensor product space? What I want to say here is that it apparently suffices to show that the tensor product space is dense in the space of smooth compactly supported functions on the cartesian product.
If anything is unclear, please let me know.
Yes, the tensor product is dense. To approximate a function $f$ in $C_c^{\infty}((a_1,b_1) \times (a_2,b_2))$ you can use functions of the form $p(x,y)\chi_1(x)\chi_2(y)$ where $p$ is any polynomial and $\chi_1,\chi_2$ are cut-off functions that are equal to $1$ on most of the interval, so that $\chi_1(x)\chi_2(y)\equiv 1$ on the support of $f$. Then the problem reduces to approximating $f$ by a polynomial in the Sobolev norm. This can be done by taking some partial derivatives of $f$, approximating them in uniform norm by Stone-Weierstrass, and integrating back: this answer by Nate Eldredge carries out the computation (in a weighted setting).