Here's another (possibly) interesting way of proving uncountability, since you tagged this as a topology question. You can inject your set into $[0,1]$ under the evaluation map

$$ (x_i) \mapsto \sum_{i = 1}^\infty \frac{x_i}{2^i} $$

This is injective on your set (but not the set of all binary sequences, since $0.0111\dots = 0.100\dots$). We shall show the image of this map is a perfect set, and hence uncountable (see Rudin, or some other source on metric spaces). Surely no point in the set is isolated, because if we have an expansion $x = 0.a_0 a_1 \dots a_2$, and if this expansion does not eventually end in all zeroes, we may truncate the expansion to find another number in the image which is as close to this number as possible. If the number does end in all zeroes, you may add a one to an arbitrary position to get another number as closed as possible. Now suppose $\lim x_i = x$, where each $x_i$ is in the image of the evaluation. We must have $0 < x < 1$, for if $x = 1$, the $x_i$ must eventually be of the form $0.11\dots$, which is impossible. Write $x = 0.a_0 a_1 \dots$. Suppose $a_k = 1$. If we force $|x_i - x| < 1/2^{k+2}$, then the first $k+2$ digits of $x_i$ must be the same as the first $k+1$ digits of $x$, hence $a_{k+1} = 0$. Thus the image is perfect, and the set is uncountable.

Intuition tells me this set is probably fractal, similar to the Cantor set, but I can't visualize it very well.