Prove or disprove: $a_{n}$ and $b_{n}$ are convergent sequences, and the sequence $c_{n}$ is defined as... 
Prove or disprove: $(a_{n})_{n\in\mathbb{N}}$ and $(b_{n})_{n \in
\mathbb{N}}\subset \mathbb{R}$ are convergent sequences and the
  sequence $(c_{n})_{n \in \mathbb{N}}$ is defined as
$$c_{n}=\left\{\begin{matrix}a_{n},\text{if n is uneven}  & \\ b_{n},
\text{if n is      even} & \end{matrix}\right.$$
The sequence $(c_{n})_{n\in\mathbb{N}}$ has got two cluster points.

I say the statement is true.
Let $a_{n}=1^{n}$ and let $b_{n}=-1^{n}$ ($a_{n}$ converges to $1$, $b_{n}$ converges to $-1$)
$\Rightarrow$ $c_{n}=(-1)^{n}$
which is a well know (divergent) sequence with two cluster points, one is at $1$ and the other one is at $-1$.
I really hope there are no mistakes?
 A: As mentioned in the comments above, as written the statement is false.  It is possible that the limits of $a_n$ and $b_n$ are the same, and thus there is exactly one cluster point, not two.
By making a modification to the statement, we can make the statement true however.  For practice and for enlightenment, we try to prove instead the statement

Given two convergent real sequences $(a_n)_{n\in\Bbb N}$ and $(b_n)_{n\in\Bbb N}$, we define the real sequence $(c_n)_{n\in\Bbb N}$ such that $c_n=\begin{cases} a_n&\text{if}~n~\text{is odd}\\ b_n&\text{if}~n~\text{is even}\end{cases}$.  Then $(c_n)_{n\in\Bbb n}$ has at most two cluster points.

Here we prove at most two as opposed to exactly two.
Let $(a_n)_{n\in\Bbb N}$ and $(b_n)_{n\in\Bbb N}$ be two convergent real sequences and let $(c_n)_{n\in\Bbb N}$ be defined as above.  As $(a_n)_{n\in\Bbb N}$ is convergent, let the limit be called $a$.  Similarly, let the limit of $(b_n)_{n\in\Bbb N}$ be called $b$.
Suppose for contradiction that there were at least three distinct cluster points.  We will call them $x,y,z$.  Now, at least one of these cluster points will be different than both $a$ and $b$.  Without loss of generality, suppose that was the cluster point $x$.
We know by definition that since $(a_n)_{n\in\Bbb N}$ is convergent to the limit $a$, for every $\epsilon>0$ there exists an $N\in\Bbb N$ such that for every $n>N$ one has $|a_n-a|<\epsilon$.  Similarly, for every $\epsilon'>0$ there exists an $N'\in\Bbb N$ such that for every $n>N'$ one has $|b_n-b|<\epsilon'$.
Set $\epsilon=\epsilon'=\frac{\min(|x-a|,|x-b|)}{2}$ and let $N$ and $N'$ satisfy the above respective definitions.  Let $M=\max(N,N')$
We have then that for every $n>M$, $|a_n-a|<\epsilon$ and $|b_n-b|<\epsilon$.
In particular we have $-\frac{|x-a|}{2}\leq-\epsilon<a_n-a<\epsilon\leq \frac{|x-a|}{2}$
By adding $a-x$ to all sides of the inequality, we have $a-x-\frac{|x-a|}{2}< a_n-x< \frac{|x-a|}{2}+a-x$.  Now, either $a-x>0$ or $a-x<0$.  In the first case, we have reading the first two parts of the inequality that $a-x-\frac{|x-a|}{2}=|a-x|-\frac{|x-a|}{2}=\frac{|x-a|}{2}<a_n-x\leq |a_n-x|$ and so $|a_n-x|>\epsilon$
In the other case, we have $\frac{|x-a|}{2}-|x-a|=-\frac{|x-a|}{2}>a_n-x$ yielding again that $|a_n-x|>\epsilon$
Analogously we can show that $|b_n-x|>\epsilon$.
As it is true for every $n\geq M$ for both $a_n$ and $b_n$, we have that $|c_n-x|>\epsilon$ for every $n\geq M$ as well.
That is to say, that for every $n\geq M$ every entry of both $a$ and $b$ will be at least $\epsilon$ away from $x$.  In particular, every odd entry of $a$ and every even entry of $b$.  However, $x$ was supposed to have been a cluster point for $c$, implying that for every $N$ (in particular $N=M$) and for every $\epsilon$ (in particular $\epsilon = \frac{\min(|x-a|,|x-b|)}{2}$) that there will be at least one $n>N$ for which $|c_n-x|<\epsilon$ but we see that no such $n$ exist.
Therefore $x$ could not have been a cluster point, implying that we made a mistaken assumption somewhere.  The only invalid assumption we made was in assuming there were at least three cluster points, therefore there can be at most two cluster points.

Perhaps a picture will help:

Now, I don't know exactly how $a_n$ and $b_n$ look, but we know that they converge, so we know that the range of possible values eventually gets less and less.  Eventually the entries of $a_n$ (pictured in red) will get arbitrarily close to the limit, so to try to draw that, I make the graph pointy like that.  Similarly for $b_n$ (pictured in blue).
The sequence $c_n$ (pictured in green) will take some values from $a_n$ and some values from $b_n$, but will always be one of these, so to picture what happens with $c_n$, all points will lie within the red and blue areas (specifically alternating back and forth between them).
The idea of the proof was to assume for contradictory purposes that there is some third cluster point other than $a$ or $b$, we call it $x$.  Wherever it appears, it will not be at the same position as $a$ or $b$ since it is unequal.  Perhaps it is "kinda close" to one of them, but if that happens, just zoom in the image and you will see a positive distance between them.
Now, if $x$ were indeed a cluster point for $c_n$, then there should be more green dots representing specific entries of $c_n$ as far to the right and as close to $x$ as we desire, however we can see from the picture that since $a_n$ and $b_n$ approach their respective limits, that they do not get closer to $x$ and will always be at least as far away from $x$ as half the distance between their limit and $x$, so there will always be some positive distance from some point ($N$) on that the values don't appear within that distance.  (The region pictured in purple)
Although this picture encompasses the main ideas (and might be convincing enough for some people), it is not a complete formal proof because we do not actually know what $a_n$ or $b_n$ look like, nor do we care.  To approach the proof then, we did as above by using the definitions and use a text-based argument.
A: (i). If $\lim_{n\to \infty}a_n=A $ and $D\ne A$ and $r\leq \frac {1}{2}|D-A|$ then the set $P=\{n:|D-a_n|<r \}$ is finite.
(ii).$\;C$ is a cluster point of $(c_n)_{\in N}$ iff for each $r>0$ the set $\{n:|C-c_n|<r\}$ is infinite.
(iii). If $\lim_{n\to \infty}a_n=A$ and  $\lim_{n\to \infty}b_n=B$ and  $c_n\in \{a_n,b_n\}$ for each $n,$ then the set of cluster points of $(c_n)_{n\in N}$ is a subset of $\{A,B\}.$
PROOF: For $C \not\in \{A,B\},$  let $r=\frac {1}{2} \min (|C-A|,|C-B|).$ Then $\{n:|C-c_n|<r\} \subset P\cup Q$ where $P=\{n:|C-a_n|<r\}$ and $Q=\{n:|C-b_n|<r\}.$ Now $P$ and $Q$ are finite by (i), so  $\{n: |C-c_n|<r\}$ is also finite.  
