As long as your goal is not proof, but just helping people get a general feel for it, there are several ways you can go.
The first is to say that, even if there were no proof, it is a worthwhile thing to try. That is, we can imagine not having the proof that it will work, but still attempting to do it anyway, just like we do with a lot of other methods. We can see that adding some numerator over $(x - c)$ and some numerator over $(x - d)$ will give us a sum that has a denominator of $(x - c)(x - d)$. We also know that it would be a heck of a lot easier to do the integral if we could find some isolated number to put in the numerator. So, why not see if there is one there.
The procedure of partial fractions is fairly straightforward if we just see if something will turn up. If we proceed just on the hunch that maybe we'll find something appropriate, then we can move forward. If we succeed we have found something we can integrate, and if we fail then, just like any other attempt, we'll have to find some other way.
In fact (I don't know if this is the case here), sometimes in math this is the way it works. You have a method that you know can work sometimes, and then, you just kind of notice that it works always. You don't know if it is "always always" or just "always for my problems", so then you search to see if you can prove it one way or the other. In any case, seeing that it could work is the first step to seeing how it works.
Now, given that it could work, it isn't too hard to see why it should always work. To see this, imagine what would happen if $(x - c)$ or $(x - d)$ had an $x$ in the numerator. When cross-multiplying, that would bring the degree up to $x^2$! Since we don't have any $x^2$ to work with, we can say for certain that either neither of them or both of them have $x$ (they could both have $x$ if one were a negative of the other to cancel them out). However, if they both had an $x$ in the numerator, then they could both be simplified through polynomial long division, which would get rid of the $x$ and leave a $1$ and a $-1$ by themselves, and a fraction without the $x$ in the numerator. When the $1$ and the $-1$ cancel out, that leaves just the fractions without the $x$ in the numerator!
So, we identified two possible cases - either both have an $x$ or neither do, and realized that the only way that both of them had an $x$ is if it were in a way that they canceled out and reduced to the case where neither one had an $x$. Thus, we can always assume that neither one has an $x$.
It's kind of convoluted, and not entirely a proof, but I think it goes a long way to showing why this method can work, and why they should trust that a proof exists.