'subscript' is unavailable: cannot subscript String with a CountableClosedRange<Int>, see the documentation comment for discussion

1. If you want to use subscripts on Strings like "palindrome"[1..<3] and "palindrome"[1...3], use these extensions.

Swift 4

extension String {
    subscript (bounds: CountableClosedRange<Int>) -> String {
        let start = index(startIndex, offsetBy: bounds.lowerBound)
        let end = index(startIndex, offsetBy: bounds.upperBound)
        return String(self[start...end])
    }

    subscript (bounds: CountableRange<Int>) -> String {
        let start = index(startIndex, offsetBy: bounds.lowerBound)
        let end = index(startIndex, offsetBy: bounds.upperBound)
        return String(self[start..<end])
    }
}

Swift 3

For Swift 3 replace with return self[start...end] and return self[start..<end].

2. Apple didn't build this into the Swift language because the definition of a 'character' depends on how the String is encoded. A character can be 8 to 64 bits, and the default is usually UTF-16. You can specify other String encodings in String.Index.

This is the documentation that Xcode error refers to.

More on String encodings like UTF-8 and UTF-16

You could just convert your string to an array of characters...

let aryChar = Array(myString)

Then you get all the array functionality...

Based on p-sun's answer

Swift 4

extension StringProtocol {
    subscript(bounds: CountableClosedRange<Int>) -> SubSequence {
        let start = index(startIndex, offsetBy: bounds.lowerBound)
        let end = index(start, offsetBy: bounds.count)
        return self[start..<end]
    }

    subscript(bounds: CountableRange<Int>) -> SubSequence {
        let start = index(startIndex, offsetBy: bounds.lowerBound)
        let end = index(start, offsetBy: bounds.count)
        return self[start..<end]
    }
}

Notable changes:

• Now an extension of StringProtocol. This allows adopters such as Substring to also gain these subscripts.
• End indices are offset from the start index of the bounds rather than the start of the string. This prevents traversing from the start of the String twice. The index method is O(n) where n is the offset from i.

Building on both p-sun's and Justin Oroz's answers, here are two extensions that protect against invalid indexes beyond the start and end of a string (these extensions also avoid rescanning the string from the beginning just to find the index at the end of the range):

extension String {

    subscript(bounds: CountableClosedRange<Int>) -> String {
        let lowerBound = max(0, bounds.lowerBound)
        guard lowerBound < self.count else { return "" }

        let upperBound = min(bounds.upperBound, self.count-1)
        guard upperBound >= 0 else { return "" }

        let i = index(startIndex, offsetBy: lowerBound)
        let j = index(i, offsetBy: upperBound-lowerBound)

        return String(self[i...j])
    }

    subscript(bounds: CountableRange<Int>) -> String {
        let lowerBound = max(0, bounds.lowerBound)
        guard lowerBound < self.count else { return "" }

        let upperBound = min(bounds.upperBound, self.count)
        guard upperBound >= 0 else { return "" }

        let i = index(startIndex, offsetBy: lowerBound)
        let j = index(i, offsetBy: upperBound-lowerBound)

        return String(self[i..<j])
    }
}

extension String {

    subscript(bounds: CountableClosedRange<Int>) -> String {
        let lowerBound = max(0, bounds.lowerBound)
        guard lowerBound < self.count else { return "" }

        let upperBound = min(bounds.upperBound, self.count-1)
        guard upperBound >= 0 else { return "" }

        let i = index(startIndex, offsetBy: lowerBound)
        let j = index(i, offsetBy: upperBound-lowerBound)

        return String(self[i...j])
    }

    subscript(bounds: CountableRange<Int>) -> String {
        let lowerBound = max(0, bounds.lowerBound)
        guard lowerBound < self.count else { return "" }

        ***let upperBound = min(bounds.upperBound, self.count-1)***
        guard upperBound >= 0 else { return "" }

        let i = index(startIndex, offsetBy: lowerBound)
        let j = index(i, offsetBy: upperBound-lowerBound)

        return String(self[i..<j])
    }
}

You getting this error because result of subscript with range is Substring? not Substring.

You must use following code:

let myString: String = "foobar"
let mySubstring: Substring? = myString[1..<3]