Use the appropriately named method String#split().

String string = "004-034556";
String[] parts = string.split("-");
String part1 = parts[0]; // 004
String part2 = parts[1]; // 034556

Note that split's argument is assumed to be a regular expression, so remember to escape special characters if necessary.

there are 12 characters with special meanings: the backslash \, the caret ^, the dollar sign $, the period or dot ., the vertical bar or pipe symbol |, the question mark ?, the asterisk or star *, the plus sign +, the opening parenthesis (, the closing parenthesis ), and the opening square bracket [, the opening curly brace {, These special characters are often called "metacharacters".

For instance, to split on a period/dot . (which means "any character" in regex), use either backslash \ to escape the individual special character like so split("\\."), or use character class [] to represent literal character(s) like so split("[.]"), or use Pattern#quote() to escape the entire string like so split(Pattern.quote(".")).

String[] parts = string.split(Pattern.quote(".")); // Split on the exact string.

To test beforehand if the string contains certain character(s), just use String#contains().

if (string.contains("-")) {
    // Split it.
} else {
    throw new IllegalArgumentException("String " + string + " does not contain -");
}

Note, this does not take a regular expression. For that, use String#matches() instead.

If you'd like to retain the split character in the resulting parts, then make use of positive lookaround. In case you want to have the split character to end up in left hand side, use positive lookbehind by prefixing ?<= group on the pattern.

String string = "004-034556";
String[] parts = string.split("(?<=-)");
String part1 = parts[0]; // 004-
String part2 = parts[1]; // 034556

In case you want to have the split character to end up in right hand side, use positive lookahead by prefixing ?= group on the pattern.

String string = "004-034556";
String[] parts = string.split("(?=-)");
String part1 = parts[0]; // 004
String part2 = parts[1]; // -034556

If you'd like to limit the number of resulting parts, then you can supply the desired number as 2nd argument of split() method.

String string = "004-034556-42";
String[] parts = string.split("-", 2);
String part1 = parts[0]; // 004
String part2 = parts[1]; // 034556-42

Sloppiness

There are several careless mistakes that are easy to fix.

Don't use implementation types in declaration, unless you really need to use a specific feature of that implementation:

ArrayList<String> arr = new ArrayList<String>();

Use the interface type instead:

List<String> arr = new ArrayList<String>();

Instead of creating a new StringBuilder, it might be faster to reset it, like this:

sb.setLength(0);

See this discussion for more details.

Don't convert a List to an array like this:

return arr.toArray(new String[0]);

It's recommended to pass a correctly sized array:

return arr.toArray(new String[arr.size()]);

Better unit tests

Give your test cases meaningful names. So that when your IDE reports some of them failing, it will be easier to understand what failed. For example:

@Test
public void testNormalSentence() {
    final String str1 = "Because";
    final String str2 = "I'm";
    final String str3 = "Batman";
    final char delim = ' ';
    String[] parts = StringUtility.split(str1 + delim + str2 + delim + str3, delim);
    Assert.assertEquals(Arrays.asList(str1, str2, str3), Arrays.asList(parts));
}

@Test
public void testStartingWithDelim() {
    final String str1 = "";
    final String str2 = "I'm";
    final String str3 = "Batman";
    final char delim = ' ';
    String[] parts = StringUtility.split(str1 + delim + str2 + delim + str3, delim);
    Assert.assertEquals(Arrays.asList(str1, str2, str3), Arrays.asList(parts));
}

@Test
public void testEmptyString() {
    final String str1 = "";
    final char delim = ' ';
    String[] parts = StringUtility.split(str1, delim);
    Assert.assertEquals(Arrays.asList(str1), Arrays.asList(parts));
}

These new names describe well exactly what they are testing. This should help understanding the code.

The assertions can be simplified, instead of:

String[] parts = StringUtility.split(str1 + delim + str2 + delim + str3, delim);
Assert.assertTrue(parts.length == 3);
Assert.assertTrue(parts[0].equals(str1));
Assert.assertTrue(parts[1].equals(str2));
Assert.assertTrue(parts[2].equals(str3));

You could simplify like this:

String[] parts = StringUtility.split(str1 + delim + str2 + delim + str3, delim);
Assert.assertEquals(Arrays.asList(str1, str2, str3), Arrays.asList(parts));

This is exactly the same, but shorter, and a lot easier to write.

I would add some more test cases:

@Test
public void testOnlyDelim() {
    final String str1 = "";
    final char delim = ' ';
    String[] parts = StringUtility.split("" + delim + delim, delim);
    Assert.assertEquals(Arrays.asList(str1, str1, str1), Arrays.asList(parts));
}

@Test
public void testNotContainingDelim() {
    final String str1 = "hello";
    final char delim = 'x';
    String[] parts = StringUtility.split(str1, delim);
    Assert.assertEquals(Arrays.asList(str1), Arrays.asList(parts));
}

Suggested implementation

@heslacher's implementation is very close to my taste. Here's a variation of that, with some minor changes:

public static String[] split(String strToSplit, char delimiter) {
    List<String> arr = new ArrayList<>();
    int foundPosition;
    int startIndex = 0;
    while ((foundPosition = strToSplit.indexOf(delimiter, startIndex)) > -1) {
        arr.add(strToSplit.substring(startIndex, foundPosition));
        startIndex = foundPosition + 1;
    }
    arr.add(strToSplit.substring(startIndex));
    return arr.toArray(new String[arr.size()]);
}

"cat".split("(?!^)")

This will produce

array ["c", "a", "t"]