Why does "x" ~~ "a", "b", "c" always give 1?

It doesn't; it evaluates to "c".

"," is a very low precedence operator. "x" ~~ "a", "b", "c" means ("x" ~~ "a"), "b", "c", which evaluates to "c" in scalar context.

And why does "x" ~~ sub { "a", "b", "c" } also give 1?

It doesn't; it evaluates to "c".

"x" ~~ sub { "a", "b", "c" } is the same as scalar( sub { "a", "b", "c" }->($x) ). Since  "a", "b", "c" evaluates to "c" in scalar context, that is what is returned.

With the ~~ operator having much higher precedence than the comma, this is parsed as a comma-list of a smart-match whose result doesn't matter, more stuff that doesn't matter at all, and "c" which is true.

The second one would pass "x" to the sub which again evaluates a comma-list and throws away everything (including the argument) but "c" which gets returned and is true.

...

Because of precedence/binding

Consider

$ perl -le " $f = 1,2,3 ; print $f"
1

$ perl -MO=Deparse -le " $f = 1,2,3 ; print $f"
BEGIN { $/ = "\n"; $\ = "\n"; }
$f = 1, '???', '???';
print $f;
-e syntax OK
[download]

Compare it to

$ perl -le " $f = ( 1,2,3 ); print $f"
3

$ perl -MO=Deparse -le " $f = ( 1,2,3 ) ; print $f"
BEGIN { $/ = "\n"; $\ = "\n"; }
$f = ('???', '???', 3);
print $f;
-e syntax OK
[download]

Since smart is binary (like = in the above example), it works on two scalars, left-scalar operator right-scalar

If you have no parentheses, the first scalar is taken, the closest, because assignment operator (=) has higher precedence than comma opeartor (it binds tighter).

If you add parenthesis, then you have the fabled List in Scalar Context, then the comma operator return the last right item (rightest most)

See Tutorials: Precedence for Idiots

Since you're confused about precedence you're probably also confused about context, so see Tutorials: Context in Perl: Context tutorial, "List" Is a Four-Letter Word, Easy Reference for Contexts

Smart match is giving me a headache...

Stop using it :) worked for me :) Smart-match