In the solution of the one-dimensional heat equation the following step is made:
Because
$$ X(x) T'(t) = \alpha X'' (x) T(t)$$
we have
$$ \frac{X''}{X} = \frac{T'}{T}$$
Why is it ok, to assume that both $X(x)$ and $T(t)$ are non-zero for all $t$ and all $x$?
That's the assumption that has to be made to do this, right?
On
If $X(x)=0$ in some dense region, then $u(x,t)=0$ in that region and we are not interested in trivial solutions. So $X(x)$ may be 0 at some points $x=x_1,x_2,\ldots$. In that case we split the initial domain into open intervals $(x_1,x_2),(x_2,x_3),\ldots$, solve the equation in each of them and then stitch the functions together. However, since we know that linear equations like $X''/X=-\lambda$ have solutions with removable singularities, we just ingore it.
Tldr. We assume that points where $X(x)=0$ or $T(t)=0$ don't cause problems and in the end we can check that it is so.
This is an important question because it is glossed over in some texts. You are assuming (hoping, guessing) that there is a product solution. Unless $X$ and $T$ are identically zero, there is a point $x_1$ and a point $t_1$ where $X(x_1) \ne 0$ and $T(t_1)\ne 0$. A basic fact about any continuous function is that if it is nonzero at some point, then it is nonzero in some neighborhood of that point. Therefore you can do the division in at least some intervals of $x$ and $t$, and find possible functions $X$ and $T$ in those intervals.
Finally you have to check your function $X(x)T(t)$ in the PDE to see whether it really is a solution. That is when you find out that your product works (or not) even for places where some factor is zero.