Make $1$

There are $\binom{10}{3} = 120$ subsets of three elements of \{1,\dots,10} . However, we can quickly reduce the number we need to check by removing:

All sets containing 1. (See the first example.)

All sets containing two consecutive integers. For instance, since $8 - 7 = 1$ , any set of the form $\{7,8,x\}$ can be used to make 1.

This leaves any subset of three elements of $\{2,\dots,10\}$ without consecutive numbers. There are only $\binom{7}{3} = 35$ of these sets. We reduce them further.

All sets of the form $\{x,y,x+y\}$ ; $\{x,y,x+y\pm 1\}$ . Examples: $\frac{2+7}9 = 1;\ \ \ \ (2 + 7) - 8 = 1;\ \ \ \ 10 - (2 + 7) = 1.$ This rules out 21 further sets: $\begin{array}{cccccc} 2,4,6 & 2,5,7 & 2,6,8 & 2,7,9 & 2,8,10 & \\ 2,4,7 & 2,5,8 & 2,6,9 & 2,7,10 & & \\ 3,5,8 & 3,6,9 & 3,7,10 & & 4,6,10 & \\ 3,5,7 & 3,6,8 & 3,7,9 & 3,8,10 & 4,6,9 & 4,7,10 \\ 3,5,9 & 3,6,10 & & & & \end{array}$

All sets of the form $\{x,y,xy\}$ ; $\{x,y,xy-1\}$ , $\{x,y,xy+1\}$ . This rules out 4 more: $(2,4,7);\ 2,4,8;\ 2,4,9;\ 2,5,9;\ 2,5,10$

All sets of the form $\{x-1,y,xy\}$ ; $\{x+1,y,xy\}$ . For instance: $6 - \frac{10}{2} = 1.$ This rules out 3 more: $3,5,10;\ 2,4,10;\ 2,6,10.$

After all this, we have 35 - 21 - 4 - 3 = $\boxed{7}$ sets left; they are: $4,6,8;\ \ 4,7,9;\ \ 4,8,10;\ \ 5,7,9;\ \ 5,7,10;\ \ 5,8,10;\ \ 6,8,10.$ It is not difficult to check that these cannot be reduced to 1 with the standard operators.