= Solution
Use <uncountability by interleaving prescribed binary coordinates>. For each arbitrary <binary sequence> $b=(b_k)_{k\geq1}$, define another <binary sequence> by
$$
x_{3k-2}=0,\qquad x_{3k-1}=a_{3k-1},\qquad x_{3k}=b_k\quad(k\geq1).
$$
There are infinitely many zero coordinates, namely $3k-2$, so $x\in X$. There are also infinitely many coordinates agreeing with the prescribed <binary expansion>, namely $3k-1$, so $x\in Y$. The coordinates $3k$ recover $b$, making this map an <injection> of the <uncountable set> of all <binary sequences> into $Y$. If $Y$ were a <countable set>, composing with an <injection> into $\mathbb N$ would make all <binary sequences> countable, contrary to part (i). \b[Therefore $\boxed{Y\text{ is uncountable}.}$] The construction works for every prescribed <binary sequence> $(a_n)$; no special distribution property of the digits of $\sqrt2$ is needed.
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