You can do this with the self-JOIN and inequality conditional expressions:

SELECT DISTINCT t0.*
FROM [table] t0
INNER JOIN [table] t1 ON t1.ID < t0.ID AND t1.TimeStamp > t0.TimeStamp

You can use the MAX instead:

select *
from (
    select *, max(timestamp) over(order by id ROWS UNBOUNDED PRECEDING) as maxVal
    from (
        VALUES  (1, N'2022-10-10')
        ,   (2, N'2022-10-11')
        ,   (3, N'2022-10-12')
        ,   (4, N'2022-10-13')
        ,   (5, N'2022-08-01')
        ,   (6, N'2022-08-02')
        ,   (7, N'2022-08-03')
        ,   (8, N'2022-10-14')
        ,   (9, N'2022-10-15')
    ) t (ID,TimeStamp)
) x
where x.maxVal > x.TimeStamp

This checks so no previous IDs are higher than the current ID and outputs:

When asking questions like this it is helpful to provide the DDL/DML. Using your example data:

DECLARE @Table TABLE (ID INT IDENTITY, TimeStamp DATE);

INSERT INTO @Table (TimeStamp) VALUES
('2022-10-10'), ('2022-10-11'), ('2022-10-12'), ('2022-10-13'),
('2022-08-01'), ('2022-08-02'), ('2022-08-03'), ('2022-10-14'),
('2022-10-15');

Using this:

SELECT *
  FROM (
        SELECT ID, TimeStamp, MAX(TimeStamp) OVER (ORDER BY ID) AS mTimeStamp
          FROM @Table
       ) a
 WHERE TimeStamp < mTimeStamp;

Here we're using a windowed function MAX to find the largest date ordered by the ID column. Using an outer query we can then compare the TimeStamp to this MAX value and determine if the current row is less than the windowed maximum.

Check for existence of an "earlier" row with a later date:

select * from T t1
where exists (
    select 1 from T t2
    where t2.id < t1.id and t2."timestamp" > t1."timestamp"
);

Your data would suggest that dates are unique but it's not clear you've described the most general problem.