Infinite compact subset of $\mathbb{Q}$ Can I find an infinite set in $(\mathbb{Q},\mathcal{T}_e|_\mathbb{Q})$ which is compact?
 A: In fact, more is known: 

Theorem: Every (Hausdorff) countable compact is homeomorphic to a subspace of $\mathbb{Q}$.

Essentially, it is a consequence of Sierpinski-Mazurkiewicz theorem, classifying (Hausdorff) countable compact topological spaces. 

For any topological space $X$, let $X'$ denote $X$ minus its isolated points. Then we define by transfinite induction:


*

*$X^{(0)}=X$,

*$X^{(\alpha+1)}= (X^{(\alpha)})'$ for any ordinal $\alpha$,

*$X^{(\lambda)}= \bigcap\limits_{\beta<\lambda} X^{(\beta)}$ for any limit ordinal $\lambda$.


Now, if $X$ a countable compact space, there exists a countable ordinal $\alpha$ such that $X^{(\alpha)}$ is finite (so $X^{(\beta)}= \emptyset$ for all $\beta > \alpha$). If $\mathrm{card} ~ X^{(\alpha)} = n$, we say that $(\alpha,n)$ is the characteristic system of $X$. Then Sierpinski-Mazurkiewicz theorem states:

Theorem: Let $\alpha$ be a countable ordinal and $n \geq 1$. Up to homeomorphism, there exists only one (Hausdorff) countable compact topological space whose characteristic system is $(\alpha,n)$.

Therefore, it is sufficient to show that $\mathbb{Q}$ contains a compact subspace whose characteristic system is $(\alpha,n)$, for every $\alpha< \omega_1$ and $n \geq 1$.
Claim 1: If $\mathbb{Q}$ contains a compact subspace $X$ whose characteristic system is $(\alpha,1)$ then it contains another such subspace whose characteristic system is $(\alpha,n)$, for every $n \geq 1$.
Let $Y= \coprod\limits_{i=1}^n X_i$ be the union of $n$ translates of $X$. Because $X$ is bounded, we may suppose that $X_i \cap X_j = \emptyset$ when $i \neq j$. Clearly, $Y$ is a compact subspace whose characteristic system is $(\alpha,n)$. $\square$
Claim 2: For very countable ordinal $\alpha$, $\mathbb{Q}$ contains a compact subspace whose characteristic system is $(\alpha,1)$. 
It can be done by transfinite induction. 
Let $X \subset \mathbb{Q} \cap (0,1)$ be compact subspace whose characteristic system is $(\alpha,1)$. Then $$Y= \coprod\limits_{n \geq 1} \left( \frac{1}{n(n+1)} X+ \frac{1}{n+1} \right) \coprod \{0\}$$ is a compact subspace whose charateristic system is $(\alpha+1,1)$.
Let $\lambda$ be a countable limit ordinal and let $\{\beta_1, \beta_2, \dots\}$ denote the ordinals smaller that $\lambda$. For every $i \geq 1$, let $X_i$ be a compact subspace of $\mathbb{Q}$ whose characteristic system is $(\beta_i,1)$; moreover, we may suppose that $X_i \subset \left( \frac{1}{i+1}, \frac{1}{i} \right)$. Then $$Y= \coprod\limits_{i \geq 1} X_i \coprod \{0\}$$ is a compact subspace whose characteristic system is $(\lambda,1)$. $\square$

Notice that Brian M. Scott gave the simplest example of infinite compact subspace: its characteristic system is $(1,1)$. But it can be generalized to get more sophisticated examples: let $$X_k= \left\{ \frac{1}{n_1} + \cdots + \frac{1}{n_k} \mid n_i \geq 1 \right\} \cup \{0\}.$$
Then $X_k$ is a compact subspace of $\mathbb{Q}$ whose characteristic system is $(k,1)$. (It is sufficient to notice that $X_k'=X_{k-1}$.)
For more information about Sierpinski-Mazurkiewicz theorem, you can see for example here.
A: $$\{0\}\cup\left\{\frac1n:n\in\Bbb Z^+\right\}$$
A: Let $a_n$ be a non constant sequence of rational numbers which converges to a rational number $a$. then $A = \{a_n\}\cup\{a\}$ is an infinite compact subset of $\mathbb{Q}$.actually we can take union of finitely many such  sets.
