EDIT: This is modified from the original version which was incorrect, as pointed out by @KajetanJaniak.

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:
$$
t=\sum_{n=0}^\infty \sum_{j=2^{2n}}^{2^{2n+1}-1}3^{-j}\mathbf 1_{j\text{ odd}}.
$$
In base 3 it's
$$
t=0.1|00|0101|00000000|0101010101010101|00000000000000000000000000000000|01\ldots
$$
(the $|$ are not part of the number, but just make it easier to read).
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 01's "01-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 01-blocks. A calculation shows that the only way to avoid having 1's in the sum in the interior of the 01-blocks is for $x$ to consist of a concatenation of 0022's and 2022's in those blocks and $x+t$ to consist of 0200's and 2200's there. Since the blocks occurring in $x+t$ have one less two than the corresponding blocks in $x$, we see that $x+t$ "looks different" on the 0-blocks than it does on the 01-blocks. Hence $x+t$ cannot have a periodic base 3 expansion, and must be irrational.