How to create a sequence with leading zeroes using brace expansion

Prefix the first number with a 0 to force each term to have the same width.

$ echo {08..10}
08 09 10

From the bash man page section on Brace Expansion:

Supplied integers may be prefixed with 0 to force each term to have the same width. When either x or y begins with a zero, the shell attempts to force all generated terms to contain the same number of digits, zero-padding where necessary.

Also note that you can use seq with the -w option to equalize width by padding with leading zeroes:

$ seq -w 8 10
08
09
10

$ seq -s " " -w 8 10
08 09 10

If you want more control, you can even specify a printf style format:

$ seq -s " " -f %02g 8 10
08 09 10

if you use printf

printf "%.2d " {8..10} 

this will force to be 2 chars and will add a leading 0. In case you need 3 digits you can change to "%.3d ".

I have the same bash version of the original poster (GNU bash, version 3.2.51(1)-release) and {08..12} doesn't work for me, but the following does:

for i in 0{8..9} {10..12}; do echo $i; done
08
09
10
11
12

It's a little more tedious, but it does work on the OP version of bash (and later versions, I would assume).