How many subsets of set $\{1,2,\ldots,10\}$ contain at least one odd integer? 
How many subsets of set $\{1,2,\ldots,10\}$ contain at least one odd integer?     

My Working:
What I can think of is subtracting the no. of subsets that don't contain a single odd number from the total number of subsets as if we calculate it for single cases (like $1$ odd integer, $2$ odd integers, $\ldots$) it would be pretty long.    
As there needs to be no odd integer, the maximum number of elements in the set is $5$ (only $5$ even integers are there in the superset).   
Case 1: $0$ elements: $1$ set
Case 2: $1$ element: $5$ sets ($1$ even integer in each set)
Case 3: $2$ elements: $(5)(5)$ sets ($1$ element odd and $1$ even) $+\binom{5}{2}$  sets (both elements even)
which gives $35$ sets     
Case 4: $3$ elements: $\cdots$
Problem:
This is getting complicated and I'm pretty sure I'll mess up if I proceed further. Is there any other way of solving this question?

Others have already explained the easy solution, here is an alternative more similar to what you tried.
We want to know how many subsets contain exactly $k$ odd integers, from $k = 1$ to $5$, and sum.


*

*$k = 1$: $\binom{5}{1} = 5$ possible subsets of $\{1,3,5,7,9\}$

*$k = 2$: $\binom{5}{2} = 10$ possible subsets of $\{1,3,5,7,9\}$

*$k = 3$: $\binom{5}{3} = 10$ possible subsets of $\{1,3,5,7,9\}$

*$k = 4$: $\binom{5}{4} = 5$ possible subsets of $\{1,3,5,7,9\}$

*$k = 5$: $\binom{5}{5} = 1$ possible subsets of $\{1,3,5,7,9\}$


In each case, we can add some even integers, so we multiply by $2^5 = 32$.
Then,
$2^5 \sum_{k=1}^5 \binom{5}{k} = 32 (5+10+10+5+1) = 992$
But effectively this can be simpler:
$2^5 \sum_{k=1}^5 \binom{5}{k} = 2^5 \left( \sum_{k=0}^5 \binom{5}{k} - \binom{5}{0} \right) = 2^5 \left( 2^5 - 1 \right) = 2^{10} - 2^5 = 992$
