Existence of a sequence distinct points for f: $\mathbb{R}^2 → \mathbb{N}$ 
Suppose that f : $\mathbb{R}^2 → \mathbb{N}$  is a function. Show that there exists a sequence of distinct points
  $x_n$ in $\mathbb{R}^2$
  so that $x_n$ → x for some x ∈ $\mathbb{R}^2$
  and f(x) = f($x_n$) for all n ∈ $\mathbb{N}$.

I'm guessing it has something to do with the cardinality of $\mathbb{R}^2$ and $\mathbb{N}$, but I'm not sure how to make a connection between this and convergence. Any hints? Thanks!
 A: The cardinality step is simply the statement that there exists $k \in \mathbb{N}$ such that $f^{-1}(k)$ is uncountable, for if each $f^{-1}(k)$ were countable then $\mathbb{R}^2 = \bigcup_{k \in \mathbb{N}} f^{-1}(k)$ would be countable, a contradiction.
But that's not enough yet, you then have to prove that the set $f^{-1}(k)$, like any uncountable subset of $\mathbb{R}^2$, has an uncountable and bounded subset $X$; again this is by contradiction, using that a countable union of countable sets is countable.
The last thing to do is to prove that $X$, like any bounded uncountable set, has a sequence of distinct points that converges to a point in $X$.
A: There is an integer $k$ such that (as already pointed out an the answer by Lee Mosher), the set $K=f^{-1}(k)$ is uncountable. 
We will show that there is a point $x\in K$ such that every open disk $B(x,\varepsilon)$ intersects $K$ in uncountably many points. 
Once this is done, we could pick $x_n\in K\cap B(x,\frac1n)\setminus\{x\}$, and then $x_n\to x$ and $f(x_n)=k=f(x)$ for all $n$. 
Note that for every $B(x,\varepsilon)$ there are rational $p,q,r$ with $r>0$ such that $x\in B((p,q),r)\subseteq B(x,\varepsilon)$. That is, there is a disk $B((p,q),r)$ such that its center $(p,q)$ has both coordinates rational, and its radius $r$ is rational, and $x$ belongs to $B((p,q),r)$ which is contained in $B(x,\varepsilon)$. (We could pick rational $p,q$ such that $(p,q)\in B(x,\frac\varepsilon3)$ and rational $r$ with $\frac\varepsilon3<r<\frac{2\varepsilon}3$.) 
In view of the above, in order to complete the proof it would be enough to show that there is a point $x\in K$ such that $B((p,q),r)\cap K$ is uncountable whenever $x\in B((p,q),r)$ with $p,q,r$ rational and $r>0$.
Suppose, towards a contradiction, that there was no such point $x\in K$. Then for every $x\in K$ we could pick rational $p_x,q_x,r_x$ with $r_x>0$ such that $x\in B((p_x,q_x),r_x)$ and $B((p_x,q_x),r_x)\cap K$ is countable. Then clearly $K=\{x:x\in K\}\subseteq \bigcup_{x\in K} \Bigl(B((p_x,q_x),r_x)\cap K\Bigr)\subseteq K$. But even though there are uncountably many $x\in K$, the balls $B((p_x,q_x),r_x)$ are only countably many (since $p_x,q_x,r_x$ are rational). Since each $B((p_x,q_x),r_x)\cap K$ is countable we obtain that $K=\bigcup_{x\in K} \Bigl(B((p_x,q_x),r_x)\cap K\Bigr)$ is the union of countably many countable sets, and hence $K$ is countable. This is a contradiction which completes the proof. 
(If you know what a topology is, note that the family $\mathcal B$ of all open disks with  rational center and rational radius is a countable base for the usual topology of the plane. We could have used any countable open base in place of $\mathcal B$. The existence of a point $x$ as above is also discussed in MSE question here and MSE question here.) 
