I don't think this should be hard. Think base 3. Something is irrational if and only if it does not have a ultimately periodic base 3 expansion. Something's in $C$ if and only if it has no 1's in its base 3 expansion.
Here's an explicit $t$ that I think does the job: $$ t=\sum_{n=0}^\infty \sum_{j=2^{2n}}^{2^{2n+1}-1}3^{-j}. $$ In base 3 it's $$ t=0.10011110000000011111111111111110000000000000000000000000000000011\ldots $$ Since $t$ does not have an ultimately periodic base 3 expansion, it's irrational.
Let's call the coordinate ranges where $t$ has 0's 0-blocks' (i.e. from $2^k$ to $2^{k+1}-1$ for $k$ odd) and the coordinate ranges where $t$ has 1's 1-blocks'.
Now if you form $x+t$ for any $x\in C$, $x+t$ has arbitrarily long blocks with no 1's in the base 3 expansion (corresponding to the blocks of 0's in $t$) - if you're unlucky, the last digit of $x+t$ in a 0-block might be a 1, but there are no others.
Hence if assume for a contradiction that $x+t$ is rational (and so has ultimately periodic base 3 expansion, the repeating block must contain only 0's and 2's.
Now consider the expansion of $x+t$ on the 1-blocks. A calculation shows that the only way to avoid having 1's in the interior of the blocks is for $x$ to have 2's in those blocks and $x+t$ to consist of 0's in the 1-blocks.
Now since $x+t$ has arbitrarily long blocks of 0's, we deduce that the periodic block must consist of all 0's -- that is $x+t$ is a triadic rational.
Since $1-t$ has ternary expansion $$ 1-t=0.122111122222222111111111111111122222222\ldots $$ it's easy to check that there's no $x\in C$ with the property that $x+t$ is a triadic rational.