bash array with square bracket strings

I can make the code you presented produce the output you wanted, or not, depending on the values I assign to variables filenames and choices.

First, I observe that bash indexed arrays are indexed starting at 0, not 1. If you are using the values 1 and 2 as indices into array filenames, and if that is an indexed array with only two elements, then it may be that ${filenames[2]} expands to nothing. This would be the result if you initialize filenames like so:

# NOT WHAT YOU WANT:
filenames=(A B)

Instead, either assign array elements individually, or add a dummy value at index 0:

# Could work:
filenames=('' A B)

Next, I'm suspicious of choices. Since you're playing with arrays, I speculate that you may have initialized choices as an array, like so:

# NOT CONSISTENT WITH YOUR LATER USAGE:
choices=(1 2)

If you expand an array-valued variable without specifying an index, it is as if you specified index 0. With the above initialization, then, $choices would expand to just 1, not 1 2 as you intend. There are two possibilities: either initialize choices as a flat string:

# Could work:
choices='1 2'

or expand it differently:

# or expand it this way:
for i in "${choices[@]}"

. Do not overlook the quotes, by the way: that particular form will expand to one word per array element, but without the quotes the array elements would be subject to word splitting and other expansions (though that's moot for the particular values you're using in this case).

The quoting applies also, in general, to your echo command: if you do not quote the expansion then you have to analyze the code much more carefully to be confident that it will do what you intend in all cases. It will be subject not only to word splitting, but pathname expansion and a few others. In your case, there is a potential for pathname expansion to be performed, depending on the names of the files in the working directory (thanks @CharlesDuffy). It is far safer to just quote.

Anyway, here is a complete demonstration incorporating your code verbatim and producing the output you want:

#!/bin/bash

filenames=('' 'A' 'B')
choices="1 2"

selections=()
for i in $choices
do
    selections+=("role[${filenames[$i]}]")
done

echo ${selections[@]}

# better:
# echo "${selections[@]}"

Output:

role[A] role[B]

Finally, as I observed in comments, there is no way that your code could output "roles", as you claim it does, given the inputs (variable values) you claim it has. If that's in fact what you see, then either it is not related to the code you presented at all, or your inputs are different than you claim.