The elementary events of your initial attempt at a sample space are triples, $\ \big(U_{i_1}, U_{i_2}, U_{i_3}\big)\ $ of units in which $\ 1\le100, 1\le i_2\le99\ $ and $\ 1\le98\ $. This isn't quite right, because it includes triples such as $\ \big(U_5, U_5, U_6 \big)\ $, in which the same unit, $\ U_5\ $, has been chosen twice, and it doesn't include triples such as $\ \big(U_1, U_{100}, U_{99}\big) $, where the second unit chosen is $\ U_{100}\ $ or the third unit chosen is $\ U_{99}\ $ or $\ U_{100}\ $. Your description of the problem, however, implies that the sampling is done without replacement—that is, the three units chosen must be distinct—and the second and third units chosen could be any one of the $100$ units.
You also need to distinguish the defective units from the non-defective ones, which you could do by choosing $5$ of your indices—$1$ to $5$, for example, or $96$ to $100$—to be those of the defective units.
In setting up the event whose probability you're looking for, you've chosen to distinguish the units by the order in which they're chosen. This isn't really necessary for this particular problem, but it's nevertheless a perfectly sound approach. If you let $\ A(1)\ $, $A(2)\ $, and $A(3)\ $ be the first, second and third units chosen, then $\ A:\{1,2,3\}\rightarrow$$\big\{U_1,U_2,\dots,U_{100}\big\}\ $ is a function which must have the property that $\ A(1), A(2), A(3)\ $ are all distinct—that is, it must be one-to-one. So you could take your sample space to be
$$
\Omega=\big\{A:\{1,2,3\}\rightarrow\big\{U_1,U_2,\dots,U_{100}\big\}\,\big|\,A\ \text{is one-to-one}\big\}\ .
$$
This sample space contains $\ 100\cdot99\cdot98\ $ elementary outcomes, all of which are equally likely, and if you take $\ U_{96}, U_{97}, \dots, U_{100}\ $ to be the defective units, then the event whose probability you're looking for is
$$
\big\{A\in\Omega\,\big|\,A(1),A(2),A(3)\in\big\{U_1,U_2,\dots,U_{95}\big\}\big\}\ .
$$
This event comprises $\ 95\cdot94\cdot93\ $ elementary outcomes, each of probability $\ \frac{1}{100\cdot99\cdot98}\ $, so its probability is $\ \frac{95\cdot94\cdot93}{100\cdot99\cdot98}\ $, as you've already found.
The reason why it's not really necessary to distinguish the order in which the units are chosen is that the occurrence or non-occurrence of the event whose probability you're interested in doesn't depend in any way on the individual values of $\ A(1), A(2), A(3)\ $ separately, but merely on whether the set of them, $\ \big\{A(1),A(2),A(3)\big\}\ $ is a subset of $\ \big\{U_1,U_2,\dots,U_{95}\big\}\ $ or not, so you can simplify your sample space a little by taking it to be the collection of $3$-element subsets of $\ \big\{U_1,U_2,\dots,U_{100}\big\}\ $:
$$
\Omega=\big\{A\subseteq \big\{U_1,U_2,\dots,U_{100}\big\}\,\big|\,|A|=3\,\big\}\ .
$$
This sample space comprises $\ {100\choose3}\ $ elementary outcomes, again all equally likely, and the event whose probability you're interested in is now
$$
\big\{A\subseteq \big\{U_1,U_2,\dots,U_{95}\big\}\,\big|\,|A|=3\,\big\}\ ,
$$
Which contains $\ {95\choose3}\ $ elementary outcomes, each of probability $\ \frac{1}{100\choose3}\ $, so its probability is $\ \frac{95\choose3}{100\choose3}\ $, with the same value as before.
Response to OP's comment
Your sample space can consist of any mathematical objects you like, as long as the model you construct is a faithful representation of the possible outcomes that can occur in reality. On reflection, given the approach you had already taken, I think my use of functions to represent the outcomes wasn't the best choice, since a small modification of your own approach would have sufficed. You can simply take your sample space $\ \Omega\ $ to be given by
$$
\Omega=\big\{\big(A_1,A_2,A_3\big)\in{\cal U}^3\,\big|\,A_1\ne A_2\ne A_3, A_1\ne A_3\big\}\ .
$$
where $\ {\cal U}= \big\{U_1,U_2,\dots,U_{100}\big\}\ $.