First part of the proof


Let a = inf(A),  b = inf(B).


Since the subsets A and B are bounded in R, the values 'a' and 'b' do exist and are defined properly.


The fact that a = inf(A) means that there is an infinite sequence  of elements  in A
which converges to 'a'.


The fact that b = inf(B) means that there is an infinite sequence  of elements  in B
which converges to 'b'.


Then the sequence    converges to value  a+b.

This simple elementary statement is easy to prove.


It implies that 

     inf(A+B) <= a+b.    (1)



            Second part of the proof


Again, let  a = inf(A),  b = inf(B).


Since the subsets A and B are bounded in R, the values 'a' and 'b' do exist and are defined properly.


Since the subsets A and B are bounded in R, the set of all real numbers of the form {x+y), 
where x is from A and y is from B, is bounded,  too.


Hence, the set of all sums (x+y) has the infinum.  Let z = inf(A+B).


The fact that z = inf(A+B) means that there is an infinite sequence  
with elements  in A  and    in B, which converges to z.


Notice that all    are not less than 'a',  and  all    are not less than 'b', 
due to the definitions of 'a' and 'b'.


It means that

     z = inf(A+B) = ( lim  as i -->  ) >= a + b = inf(A) + inf(B).    (2).


Inequalities (1) and (2), taken together,  prove that

    inf(A+B) = inf(A) + inf(B).


QED.