Given a set s of m numbers {1,2,3,4,....m} i need to prove that given any number x in the interval [1,m(m+1)/2] i can select a subset s' from the set s such that sum of elements in s' is equal to x.  

My approach was to consider two cases:  
1) 1<=x<=m  
In this case s' is just a singleton set containing x

2)m<x<= (m\*(m+1))/2  

In this case i first find the greatest integer 'i' such that x - [m+(m-1)+(m-2)+(m-3).....(m-i)] is positive. Let this quantity be c = x - [m+(m-1)+(m-2)+(m-3).....(m-i)] . Now because of the way we have defined 'i' , if we subtract m-i-1 from c we will get a non positive value.   
 This means that c-(m-i-1)<=0.  
This means that c<=(m-i-1).
Also 0<=c.   

if c=0 we have,  

c = x - [m+(m-1)+(m-2)+(m-3).....(m-i)] = 0.  
This means x = m+(m-1)+(m-2)+(m-3).....(m-i)  
And we can select s' containing elements m,(m-1),(m-2),(m-3).....(m-i).  

but if c is not equal to 0,  

Then we know that 1<=c<=(m-i-1), We can select this value of c.  
This means we can select s' as m,(m-1),(m-2),(m-3).....(m-i), c   

Is this proof correct??  

How can this be proved using induction??