Combi-natrics!

Consider the 5 pairs of elements of $A$ which sum to $11$ , namely $(1,10), (2,9), (3,8), (4,7)$ and $(5,6)$ . The desired subsets of $A$ cannot contain both elements of any of these pairings, but can contain one or the other, or neither, of the elements of each pairing. This leaves us with 3 options for each of the 5 pairings, resulting in a total of $3^{5} = \boxed{243}$ subsets of $A$ in which no two elements sum to $11$ .

We can rephrase the question as selecting $1$ from each of the following

A) $1|10|\phi$

B) $2|9|\phi$

C) $3|8|\phi$

D) $4|7|\phi$

E) $5|6|\phi$

Result is $3$ from each case. Using the fundamental theorem of counting (multiplication), we get the answer as

$3^{5}=\boxed {243}$