Let Xn be a bounded set of integers. Show that Xn has a subsequence that is eventually constant. This is an extra unassigned problem for an introductory real analysis class that I am currently taking. That being said, I haven't been able to find a hint or a suggestion elsewhere. So far this is what I have.

Let $X_n$ be a bounded set of integers with a subsequence that is
  eventually constant.
By the Bolzano-Weierstrass Theorem, $X_n$ contains a convergent
  subsequence, $(X_n)_k$.
Because $(X_n)_k$ is convergent, it is eventually constant.
Let $x$ be the convergence point of $(X_n)_k$.
Then, $\exists N \in \mathbb{N}$ such that $(X_n)_k = x$ and $|(X_n)_k - N| \lt \epsilon,  \forall n \geq N$.
Thus, we can conclude that $X_n$ contains a subsequence that is
  eventually constant.

If anyone could correct this for me and give me advice as to how I can improve this proof it would be greatly appreciated. Thanks in advance.
 A: Rather than improving your proof, let me prove something more general.
A bounded set of integers is finite. So here's a general fact: for any finite set $A$, every sequence $(X_n)$ in $A$ has a constant subsequence.
The proof uses the pigeonhole principle. The sequence $(X_n)$ assigns a value $X_n \in A$ to each natural number $n$. There are infinitely many natural numbers, but only finitely many values in $A$. It follows that one of those values, call it $a \in A$, is equal to $X_n$ for infinitely many natural numbers $n$. List those natural numbers in increasing order: $1 \le n_1 < n_2 < n_3 < \cdots$. And there you have your constant subsequence: $X_{n_i}=a$ for all $i=1,2,3,\ldots$.
Notice: there's no analysis in this proof, no $\epsilon$'s or anything. It's just set theory.
A: You seem to have most the necessary bits in your proof, but they are very much out of order. As far as the proof itself, you want to start simply by assuming an example of what you are given:

Let $\{X_n\}_{n\in\mathbb N}$ be a bounded sequence of integers.

Note that we definitely cannot let this be a bounded sequence of integers containing that is eventually constant - we are only given bounded and are trying to prove that it's eventually constant, so we may not assume that, as you do in your proof. The next line is good: let's apply something we know:

By the Bolzano-Weierstrauss theorem, there is some subsequence $X_{n_k}$ that converges.

The next line should be removed; it is again assuming the result. But, then you're back on track by naming the convergence point:

Let $x$ be the convergence point of $X_{n_k}$.

Finally, you have a useful statement that you don't quite use properly: You define a limit, but what that really lets you do is substitute in any $\varepsilon>0$ and get a corresponding $N$ - critically, your proof does not do this, and thus misses an important step. Here, you want to show that these terms are constant knowing that they are integers. Let's pick $\varepsilon = 1/2$ and then write:

By definition of a limit, there is some $N\in \mathbb N$ such that if $k\geq N$ then $|X_{n_k} - x | < 1/2$.

Finally, you should proceed to the formal statement of a result, which never occurs in your proof.

Therefore, if $k, k'\geq N$ we have $|X_{n_k} - X_{n_{k'}}| < 1$ by the triangle inequality. Since these quantities are integers, it must be that $X_{n_k} = X_{n_{k'}}$. Therefore, the subsequence $X_{n_k}$ is eventually constant.

These last two steps are really important because they are what distinguishes the integers from other sets - every integer is some distance from every other one.
A: If a sequence of integer $(x_n)$ is not eventually constant, then no matter how large $N$ is, there is a $k_N>N$ such that $|x_N-x_{k_N}| \geq 1$, so $(x_n)$ cannot be convergent.
