First of all,

my $c = &func() ? say "$c found!" : say "None!";
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means

my $c = ( &func() ? say "$c found!" : say "None!" );
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Neither my $c nor the assignment has been evaluated when you tried to read $c. You meant

( my $c = &func() ) ? say "$c found!" : say "None!";
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Fixing that error won't fix the error you see, though. Lexical variables only become visible the statement after the one in which they are created in order to allow the following to work.

my $x = 123;
{
   my $x = $x;
   say $x;   # 123
   $x = 456;
   say $x;   # 456
}
say $x;   # 123
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---

Anyway,

my $c = func();
say $c ? "$c found!" : "None!";
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is much clearer than the following cleaned-up but still non-functional version of your code:

say( ( my $c = func() ) ? "$c found!" : "None!" );
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