Anthony, I have some doubts about your explicit example of $t$. Let's write it like that: $$ t = 0.1(00)^{2^0}(11)^{2^1}(00)^{2^2}(11)^{2^3}\ldots (00)^{2^{2n}}(11)^{2^{2n+1}}\ldots $$ Then take $x \in C$: $$ x = 0.0(20)^{2^0}(02)^{2^1}(20)^{2^2}(02)^{2^3}\ldots (20)^{2^{2n}}(02)^{2^{2n+1}}\ldots $$ As a result we get $$ x + t = 0.1(20)^\infty \in \mathbb Q, $$ because within each block of length $2^n$ we can calculate the sum separately. (continued)

More precisely, for each subsequent pair of digits we use the following two formulas: $$ 0.(0)^n00 + 0.(0)^n20 = 0.(0)^n20 $$ and $$ 0.(0)^n11 + 0.(0)^n02 = 0.(0)^n20. $$