Why doesn't the separator from IFS work for array expansion?

From POSIX, regarding $*:

When the expansion occurs in a context where field splitting will not be performed, the initial fields shall be joined to form a single field with the value of each parameter separated by the first character of the IFS variable if IFS contains at least one character, or separated by a <space> if IFS is unset, or with no separation if IFS is set to a null string.

To join words with a separator, you will have to use $*, or ${array[*]} in bash:

$ set -- word1 word2 word3 "some other thing" word4
$ IFS='|'
$ echo "$*"
word1|word2|word3|some other thing|word4

Or, with an array in bash:

$ arr=( word1 word2 word3 "some other thing" word4 )
$ IFS='|'
$ echo "${arr[*]}"
word1|word2|word3|some other thing|word4

With your code:

$ myarr=( 1 2 3 )
$ echo "$( IFS="|"; echo "${myarr[*]}" )"
1|2|3

Compare:

$ myarr=(1 2 3)
$ printf '%s\n' $( IFS="|"; echo "${myarr[@]}" )
1
2
3
$ printf '%s\n' $( IFS="|"; echo "${myarr[*]}" )
1|2|3

From man bash:

@
… When the expansion occurs within double quotes, each parameter expands to a separate word. That is, "$@" is equivalent to "$1" "$2" ...

*
… When the expansion occurs within double quotes, it expands to a single word with the value of each parameter separated by the first character of the IFS special variable. That is, "$*" is equivalent to "$1c$2c...", where c is the first character of the value of the IFS variable.

The description above is for positional parameters but it also apply to array expansions.