why this is not a closure though ?

Because it is just a bare block. See perlsub and perlref, section Making References and the example there for how closures work.

in my eyes my $a is visible outside the {} many thanks

Outside the scope of that bare block, you see the global $a, which is special (see sort). How can you tell that it is the same as the $a declared with my inside the block? The inner is just declared and not used in any way. Here's how to check that:

use warnings;
use strict;
$a=1;
{
    my $a = 5;
}
print $a;
__END__
1
[download]

One last thing - be sure that your example code actually compiles. Your first example doesn't, and in your second example there's a ';' missing at the end of $a=1

_($_=" "x(1<<5)."?\n".q·/)Oo.  G°\        /
                              /\_¯/(q    /
----------------------------  \__(m.====·.(_("always off the crowd"))."·
");sub _{s./.($e="'Itrs `mnsgdq Gdbj O`qkdq")=~y/"-y/#-z/;$e.e && print}