You don't need to use the set $T$ to prove continuity. It provides one way of going about it (and a handy way of thinking about limits of $t$), but it is by no means necessary. Let's do a direct approach, which works in much the same way as the hint in the linked question.
First, consider the case $\varepsilon = \frac{1}{n}$ for some $n$ (yes, $\varepsilon > 0$ should be completely arbitrary, but we'll get to that in a second). Following the hint in the linked question, we just need to show that
$$\frac{p}{q} \in \left(1 - \frac{1}{n}, 1 + \frac{1}{n}\right) \setminus \{1\} \implies |q| \ge n + 1 > n.$$
Let's say we have such a $p/q$. Then
$$\left|\frac{p}{q} - 1\right| < \frac{1}{n} \implies\frac{|p - q|}{|q|} < \frac{1}{n} \implies |q| > |p - q|n.$$
Since $p/q \neq 1$, we have $p \neq q$, hence $|p - q| \ge 1$, hence
$$|q| > |p - q|n \ge n \implies |q| \ge n + 1.$$
The benefit here is that it means
$$t\left(\frac{p}{q}\right) \le \frac{1}{n + 1} < \frac{1}{n}.$$
That is, we can simply take $\delta = \frac{1}{n}$ as well, since
$$0 < |x - 1| < \delta \implies \begin{cases}t(x) = 0 < \dfrac{1}{n} & \text{if } x \in \Bbb{R} \setminus \Bbb{Q} \\ 0 \le t(x) < \dfrac{1}{n} & \text{if } x \in \Bbb{Q}\end{cases} \implies |t(x) - 0| < \frac{1}{n},$$
proving the limit for $\varepsilon = \frac{1}{n}$.
How can we show this for more general $\varepsilon > 0$? Well, since $\frac{1}{n} \to 0$ as $n \to \infty$, we can always choose a sufficiently large $n$ such that $0 < \frac{1}{n} \le \varepsilon$. We can then use $\delta = \frac{1}{n}$ as above.
If you wanted to go about this question using the hint, let's prove the isolation of $T$. You can prove isolation directly, but here's an indirect proof.
Suppose, for the sake of contradiction $(x_n)$ is an injective sequence in $T$ such that $x_n \to x_0 \in T$. Then $x$ and $x_n$ must be rational numbers; let $x_0 = p_0/q_0$ and $x_n = p_n/q_n$, each in lowest terms, each with positive denominator. We also must have $t(x_i) = 1 / q_i < \varepsilon$, i.e. $0 < q_i < 1/\varepsilon$.
Note that there are only finitely many integers in $(0, 1/\varepsilon)$, so there must be one such integer $q$ which is used infinitely often in the sequence. That is, there must be a subsequence $x_{n_k}$ such that $q_{n_k} = q$ for all $k$. But then
$$\frac{p_{n_k}}{q_{n_k}} = \frac{p_{n_k}}{q} \to x_0 \implies p_{n_k} \to x_0 q,$$
as $k \to \infty$. This implies $p_{n_k}$ is Cauchy, and so eventually the terms must become closer than $1/2$ to each other. But since this sequence is composed of integers, this implies that $p_{n_k}$ is eventually constant, contradicting injectivity. Thus, each point in $T$ is isolated.
How does this help us prove the limit? Well, for $0 < \varepsilon \le 1$, we have $1 \in T$, as $t(1) = 1 \ge \varepsilon$. Therefore, as proven above, it is isolated. That is, there must exist some $\delta > 0$ such that $(1 - \delta, 1 + \delta) \cap T = \{1\}$. That is,
\begin{align*}
0 < |x - 1| < \delta &\implies x \in (1 - \delta, 1 + \delta) \text{ and } x \neq 1 \\
&\implies x \in (1 - \delta, 1 + \delta) \text{ and } x \notin T \cap (1 - \delta, 1 + \delta) \\
&\implies x \notin T \\
&\implies t(x) < \varepsilon.
\end{align*}
