Which sets of sequence is countable and Uncountable. Consider the sequences 

$$\displaystyle X=\left\{(x_n): x_n \in \left\{0,1\right\},n \in \mathbb{N} \right\}$$
$$and$$
$$\displaystyle Y=\left\{(x_n)\in X:x_n=1 \;\;\text{for at most finitely many n} \right\}$$

I have to choose which is uncountable and which is countable.
Solution i tried- Here $X$ is set of sequence with enteries from $\left\{0,1\right\}$  thus it has number of elements $2^{\aleph_0}$ which is uncountable .
Now The set $Y$ it has all the sequences from the set $X$ but some of its elements of sequences is replaced by the only '$1$' so its Cardinality will be less then $2^{\aleph_0}$ ,but by $\textbf{ continuum hypothesis}$ there is no set having Cardinality in-between the ${\aleph_0}$ and $2^{\aleph_0}$ so the set $Y$ will be countable 
I write this proof but i don't even know this is correct or not but i am sure about set $X$ but not sure about $Y$ please help me with set $Y$ 
Thank you.

Let $A_n=\{1,\cdots,n\}$. For each  $f \in \{0,1\}^{A_n}$, define $g_f:\Bbb N \to \{0,1\}$ by $$g_f(x)=\begin{cases}f(x)&\text{if}\;x \in \{1,2,..,n\}\\0&\text{otherwise} \end{cases}$$ Then each $g_f \in Y$. Then $Y$ can be written as  $$Y=\cup_{n=1}^\infty Y_n$$ where $Y_n=\left\{g_f: f \in \{0,1\}^{A_n}\right\}$ . Here each $Y_n$  is finite and hence $Y$ is countable!
