Please evaluate the following, and help complete the proof.
Let $R$ be the set of polynomials with rational coefficients, then $$R=\{P(x)|a_nx^n+a_{n-1}x^{n-1}+...+a_1x+a_0, a_i\in \mathbb{Q}\}$$
Given that $|R|=|\mathbb{Q}\setminus \{0\}×\mathbb{Q}_1×\mathbb{Q}_2×...×\mathbb{Q}_{n-1}×\mathbb{Q}_n|$. Whose cardinality is the same as $\mathbb{N}^{n+1}$, $R$ is countable.
Now, let $\mathbb{A}$ be the set of Algebraic Numbers, then $\mathbb{A}=\{x|P^n(x)=0, n\in[',k']\}$. Furthermore, we can denote each set of polynomial in $R$ as $$P_1=\{P'|a_1x+a_0\}, P_2=\{P''|a_2x^2+a_1x+a_0\},..., P_k=\{P^{k'}|a_kx^k+a_{k-1}x^{k-1}+...+a_1x+a_o\}$$ where each index $1-k$ corresponds to the degree(s) of the polynomial. Furthermore, for each $i\in[1,k]$ we can conclude from the first part that $|P_i|\leq |\mathbb{Q}^{n+1}|$ since a subset can not be strictly greater than the superset it belongs to. Moreover, we can also deduce that for each polynomial in $P_i$ we have $\{x_1,x_1,...,x_1^i\}$ as one of the solutions followed by $\{x_2,x_2,...x_2^i\}$, and going up to $$\{x_{r\leq |P_i|}, x_{r\leq |P_i|},..., x_{r\leq |P_i|}^i\}$$
Edit
Now, from what was said above we can deduce $$|\{x_1^i...\} \cup \{x_2^i... \} \cup \{x_{r\leq |P_i|}^i...\}|\leq |\mathbb{N}_1×\mathbb{N}_2×...×\mathbb{N}_{r\leq |P_i|}|$$
Similarly, since $|R|\leq |\mathbb{N}^{n+1}|$ we can deduce $$|\mathbb{A}| \leq|\mathbb{N}^{n+1}×\mathbb{N}^{r\leq|P_i|}|$$
Therefore, $\mathbb{A}$ is countable.
Now what can I do to extend and complete this proof? I have a strong inkling I am going the right way without constructing a bijection. I want to keep it that way.
Edit
Thanks in advance.
There are countably many polynomials in $\Bbb Z[x]$. That is because each such polynomial corresponds to a sequence of integers of finite length. (There are uncountably many formal power series with coefficients in $\Bbb Z$.) It's fairly easy to see that there are countably many sequences of any fixed length $n$. A countable collection of countably many sequences is likewise countable.
Each polynomial has only finitely many roots so the collection of roots of polynomials in $\Bbb Z[x]$ is likewise countable. But every algebraic number is a root of a polynomial in $\Bbb Z [x]$, so we are done.