You're right, the set of all rational sequences convergent to zero is uncountable. Your proof is good, here's another using Cantor's (other) diagonal argument.
Suppose they were enumerable, denote the sequences as $s_1, s_2, \dots$. Arrange these top to bottom, where $s_{i, j}$ is the $j$th element of the $i$th sequence. Now consider $s_{n, n}$. Let $k_n$ be the sequence defined by
$$k_n =
\begin{cases}
1/n & s_{n, n} \neq 1/n\\
1/(n+1) & s_{n, n} = 1/n
\end{cases}
$$
then $k_n \to 0$ but $k_n$ is not in the enumeration.
Your idea for the solution is correct, though your notation is a bit confusing. Why not write simply $S = \{(a_1, a_2, \dots) \mid 0 = a_j = a_{j+1} = \dots \text{ for some } j\}$? This is an equally valid and more intuitive explanation of the set.
Proof of denseness: fix $\varepsilon > 0$ and an arbitrary sequence $x$. For some large $N$, we have $|x_n| < \varepsilon$. Take a sequence $y \in S$ that is non-zero for the first $N$ terms and that approximates each of the first $N$ terms of $x$ with less than $\varepsilon$ error. Then $d(x, y) < \varepsilon$.
Proof of countable: $\mathbb{Q}^n$ is countable as it has the same cardinality as the disjoint union of $n$ copies of $\mathbb{Q}$. $S$ is countable since it has the same cardinality as the countable union of $\mathbb{Q}^n$ for each $n \geq 0$.
I don't understand how the last sentence of your argument is supposed to work -- how do you clam that the set you're talking about injects into $\bigcup_i \bar Q_i$? -- but what you're trying to prove is certainly a falsehood.
You know that there are uncountably many infinite binary sequences; however, if $(b_n)$ is a infinite binary sequence, then
$$ c_n = n + b_n $$
is an infinite rational sequence that diverges to $+\infty$, and the mapping from $(b_n)$s to $(c_n)$s is injective, so even just there you have uncountably many different rational sequences that diverge to infinity.
Best Answer
Note that since $\mathbb Q$ is countable we may rewrite the question in terms of sequences over $\mathbb N$ and additionally it is then easy to give a bijection between the first set you mention and the set $\mathbb{N}^\ast$ of finite tuples over $\mathbb N$ by simply mapping a sequence $(x_1, ..., x_N, c, ...)$ to the tuple $(x_1, ..., x_N, c)$.
Now, to show that $\mathbb N^\ast$ is countable one can simply note that it is the countable union of countable sets.