Find a sequence $\{x_n\}$ such that for any $y \in \Bbb R$, there exists a subsequence $\{x_{n_i} \}$ converging to $y$. Find a sequence $\{x_n\}$ such that for any $y \in \Bbb R$, there exists a subsequence $\{x_{n_i} \}$ converging to $y$.
I actually have almost zero clue where to start? Thanks to anyone who can help! 
 A: An enumeration of $\mathbb Q$.
A: As pointed out by Surd, an enumeration of $\Bbb Q$ (that means, a listing of the elements of $\Bbb Q$ - i.e. a sequence in $\Bbb Q$ ). 
The reason for this being as follows:
Since $\Bbb Q$ is dense in $\Bbb R$, we know that for any $y \in \Bbb R$ there exists a sequence $\{q_n\}\in \Bbb Q$ such that $q_n \to y$. Since the sequence $\{q_n\}$ converges to $y\in \Bbb R$, we then (clearly) must have that every subsequence $\{q_{n_k}\}$ of $\{q_n \} \in \Bbb Q$ converges to the same $y \in \Bbb R$.
A: Think about the fact that every real number can be arbitrarily approximated by a finite decimal expansion (i.e., a rational number)...
In particular, the density of the rational in the reals is a fundamental, elementary notion from real analysis. Also that the rationals are countable...
This question was similar Every real number is the limit of some subsequence of a given sequence in the sequence of all rationals
A: Although it's not necessary, it's easier if every rational occurs in the sequence infinitely often. There is a bijection $$f\colon \mathbb{N} \to \mathbb{N} \times \mathbb{Q} \text{.}$$ So let the sequence be $$q_n = \text{2nd coordinate of }f(n)\text{.}$$
Given a real $x$, let its decimal expansion be $\Sigma_{i=0}^{\infty}a_i 10^{-i}$. Then each partial sum $x_n = \Sigma_{i=0}^n a_i 10^{-i}$ occurs infinitely often in the sequence $(q_n)$, and we can define the following subsequence:
$$
\begin{align}
n_0 &= \text{least $n$ where $q_n = x_0$,} \\
n_{k+1} &= \text{least $n > n_k$ where $q_n = x_{k+1}$.}
\end{align}
$$
Clearly, $(q_{n_k})_{k\ge 0}$ is a sequence of the partial sums, in order, and $\lim_{k\to \infty} q_{n_k} = x$.
