# Find unique number of days

with cte as
(Select id, start_day as day
group by id, start_day
union
Select id, end_day as day
group by id, end_day
)

select id, count(day)
from cte
group by id


One possible approach to simplify the statement in the question(fiddle), is to use VALUES table value constructor and appropriate joins:

SELECT
t.emp_id,
SUM(CASE
WHEN d1.day_no = d2.day_no THEN 1
ELSE d2.day_no - d1.day_no
END) AS no_of_days
FROM times t
JOIN (VALUES ('monday', 1), ('tuesday', 2), ('wednesday', 3), ('thursday', 4), ('friday', 5)) d1 (day, day_no)
ON t.start_day = d1.day
JOIN (VALUES ('monday', 1), ('tuesday', 2), ('wednesday', 3), ('thursday', 4), ('friday', 5)) d2 (day, day_no)
ON t.end_day = d2.day
GROUP BY t.emp_id


But if you want to count the distinct days, the statement is different. You need to find all days between the start_day and end_day range and count the distinct days:

;WITH daysCTE (day, day_no) AS (
SELECT 'monday', 1 UNION ALL
SELECT 'tuesday', 2 UNION ALL
SELECT 'wednesday', 3 UNION ALL
SELECT 'thursday', 4 UNION ALL
SELECT 'friday', 5
)
SELECT t.emp_id, COUNT(DISTINCT d3.day_no)
FROM times t
JOIN daysCTE d1 ON t.start_day = d1.day
JOIN daysCTE d2 ON t.end_day = d2.day
JOIN daysCTE d3 ON d3.day_no BETWEEN d1.day_no AND d2.day_no
GROUP BY t.emp_id


You need to basically find the intersection of the days worked by each emp_id on each task with all the days of the week, and then count the distinct days:

with days_num as (
SELECT *
FROM (
VALUES ('monday', 1), ('tuesday', 2), ('wednesday', 3), ('thursday', 4), ('friday', 5)
) AS d (day, day_no)
),
emp_day_nums as (
select emp_id, d1.day_no AS start_day_no, d2.day_no AS end_day_no
from times t
join days_num d1 on d1.day = t.start_day
join days_num d2 on d2.day = t.end_day
)
select emp_id, count(distinct d.day_no) AS distinct_days
from emp_day_nums e
join days_num d on d.day_no between e.start_day_no and e.end_day_no
group by emp_id


Output:

emp_id  distinct_days
1       4
2       5


Demo on SQLFiddle