Succession divided into a countable infinity of subsequences Proves or Disproves: 
The terms of a sequence of real numbers are divisible into a countable infinity of subsequences (ie each element of the original sequence appears in one and only one of the subsequences), each of which converges to 1. Then also the original sequence converges to 1.
 A: Hint: Try to find countably many sequences, each with values only $0$ and $1$ and somehow mash them together. It may be convenient to use the fact that there are infinetly many primes.
Update: Enumerate the primes. For the $k$-th prime $p_k$, define a sequence $(q^{(k)}_n)_{n ∈ ℕ}$ by $q^{(k)}_n = 0$ if $n ≤ k$ and $q^{(k)}_n = 1$ if $n > k$. For every prime this sequence converges to $1$.
Now define a sequence $(a_n)_{n ∈ ℕ}$ by $a_n = q^{(k)}_N$ if $n = {p_k}^N$ for some $k ∈ ℕ$ and $a_n = 1$ otherwise.
That is on the prime powers of the $k$-th prime it is defined as above.
The sequence looks like this (starting with $n=1$):
$$1,0,0,1,0,1,0,1,0,1,0,1,0,1,1,1,0,1,0,1,1,1,0, \ldots$$
Okay, somehow I thought the pattern would become clearer by this.
Anyway, for $n=1,6,10,12,14,16,\ldots$ the sequence is constantly $1$ since the indices are no prime powers. This is a subsequence converging to $1$.
For $n=2,4,8,16,\ldots$ the sequence is $0,1,1,1\ldots$, a subsequence converging to $1$.
For $n=3,9,27,81,\ldots$ the sequence is $0,0,1,1,\ldots$, a subsequence converging to $1$.
But for any $N ∈ ℕ$ there is $p_N > N$, so $\lvert a_{p_N} - 1\rvert = \lvert 0 - 1 \rvert  = 1 > \tfrac{1}{2}$, since $N ≤ N$ and therefore $a_{p_N} = 0$.
The sequence $(a_n)_{n ∈ ℕ}$ doesn't converge.
