This fails because the open() makes STDERR go to where STDOUT was going at the time of the open(). The backticks then make STDOUT go to a string, but don't change STDERR (which still goes to the old STDOUT).

So don't use backticks. Redirect STDOUT to a file, redirect STDERR to STDOUT and use system() instead.

use strict;
use warnings;
use File::Temp;

my $temp_stdout = File::Temp->new;

local *OLDOUT;
local *OLDERR;

open( OLDOUT, ">&STDOUT" );
open( OLDERR, ">&STDERR" );
open( STDOUT, ">$temp_stdout" );
open( STDERR, ">&STDOUT" );

# Funky quoting for Windows.  Sigh.
system('perl -e "print q{to stdout}; warn q{to stderr}; print q{more t
+o stdout}');
 
close(STDOUT); 
open(STDOUT, ">&OLDOUT");
open(STDERR, ">&OLDERR");

open CAPTURED, "<$temp_stdout";
my $capture = do { local $/; <CAPTURED> };
close CAPTURED;

print "Got this:\n$capture";
[download]

That still doesn't solve the problem of keeping them in sync because the subprocess still has two buffered handles. The fact that they go to the same place doesn't matter. You need to get the child process to turn off buffering.