Remove an item in an array if the item has a duplicate.

I ran your code and looked at the contents of the unique array each time it got called:

[]
[ 'John' ]
[ 'John', 'Bean' ]
[ 'John', 'Bean' ]
[ 'John', 'Bean', 'Mike' ]
[ 'John', 'Bean', 'Mike', 'Rose' ]
[ 'John', 'Bean', 'Mike', 'Rose' ]

So it starts off empty, then it adds John, and Bean. It sees the 2nd John, ignores it, adds Mike, adds Rose, sees the last John, ignores it. You might build in some way of deleting the 1st John when it sees the 2nd John, but it somehow needs to remember that when it sees the last John to not add that one either.

My solution would be to make a helper function that first reduces the list into an object that has how many of each name there are. Then you can iterate over the names in the object, and make a new list were those names have only one occurrence:

let array = [ 'John', 'Bean', 'John', 'Mike', 'Rose', 'John'];

const countNames = (list) => {
    return list.reduce((acc, curr) => {
        if (acc[curr]) {
            acc[curr] += 1;
        } else {
            acc[curr] = 1;
        }
        return acc;
    }, {})
};

const removeDuplicates = (array) => {
    const nameCounts = countNames(array);  // { John: 3, Bean: 1, Mike: 1, Rose: 1 }
    return Object.keys(nameCounts)
        .filter(name => nameCounts[name] === 1);
};

console.log(removeDuplicates(array)); // [ 'Bean', 'Mike', 'Rose' ]

This might seem like a more complex solution, but we basically end up iterating through the list once to count everything, and then iterating through the names a second time to filter out any with a count more than one. It reduces down to Big O(n) if you're into algorithm complexity.