Q: Why this java regex returning false? My intention is to locate all the "tokens" and quoted strings with a String::split() method:
(("[\w\s]*")|(\w*))*

Input text: 
this  "a test" abd "and more"

It always returns false. 
Also, how do I "quote" it so I can use it in the source code?

A: String.split splits using the matches as the separators, i.e. it returns everything that doesn't match. If you want to return what you're matching, you should use Matcher.find.
Also, \\w* matches a length 0 string, which you probably want to avoid. Use + to match 1-or-more.
You also have some unnecessary brackets and the outer * should not be there - the rest of the regex already matches all characters in a single token and multiple tokens should presumably be matched across multiple invocations of the regex matching (so saying "any number of these" doesn't make sense).
Matcher m = Pattern.compile("\"[\\w\\s]*\"|\\w+").matcher("this  \"a test\" abd \"and more\"");
while (m.find())
    System.out.println(m.group());

The above prints:
this
"a test"
abd
"and more"

To remove the quotes, you can update the regex to use look-around, which would check if the "'s are there, but won't actually match them:
"(?<=\")\\w[\\w\\s]*(?=\")|\\w+"


To understand how to escape things, you need to keep the different layers in mind. First there's Java itself, where having a " would start or end a string, so that needs to be escaped with a \ if you want the " character to appear in a string. Then there's the regex code, which expects \w and \s, but Java doesn't allow \ without it being escaped, so that's \\w and \\s.
