Sequences and Intervals I came across another real analysis problem in my self study:

Let $[a,b]$ be a closed interval in $\mathbb{R}$ and let $(x_n)$ be any sequence in $\mathbb{R}$. Prove that $[a,b]$ contains a real number not equal to any term of the sequence.  

I think I need to use the nested interval theorem:
Theorem. If $(I_n)$ is a nested sequence of closed intervals, then the intersection of the $I_n$ is nonempty. In other words, if $I_n = [a_n, b_n]$, where $a_n \leq b_n$ and $I_1 \supset I_2 \supset I_3 \supset \dots$ and $a = \sup \{a_n: n \in \mathbb{Z}^{+} \}$, $b = \inf \{b_n: n \in \mathbb{Z}^{+} \}$ then $a \leq b$ and $\bigcap_{n=1}^{\infty} [a_n, b_n] = [a,b]$. 
It seems obvious if we know that the interval is uncountable and the sequence is countable. Or could you do the following: Pick an arbitrary element $x_0$ of $(x_n)$ in $[a,b]$ (if there is none then we are done). By denseness, there is a real number $\alpha$ between $a$ and $x_0$. If $\alpha$ is in the sequence pick another number $\alpha_1$ between $a$ and $\alpha$. Keep doing this until you find a number not in the sequence. 
Would this idea work?
 A: To use the nested interval theorem, you divide the interval in three parts (compact intervals of equal length). There is a part $I_1$ which does not contain $x_1$. Next, divide this part in three parts. There is one part $I_2 \subset I_1$ which doesn't contain $x_2$ or $x_1$. Doing this inductively you get a decreasing sequence of intervals $(I_n)$ such that $I_n$ does not contain $x_1,...,x_n$. The intersection of these intervals is nonvoid, and it does not contain any of the elements of the sequence.
A: The interval [a,b], as a closed subset of R, is a complete metric space . If the union of the terms of ${a_n}$ contained [a,b], that would violate Baire Category Theorem, which states that a complete metric space cannot be the countable union of nowhere-dense sets; any singleton is nowhere-dense in [a,b]; actually, the sequence as a whole--even if it converges--is nowhere-dense in [a,b].  
Edit: My answer above was a misuse (or unhelpful use) of big machinery.
Consider these two cases:
1) ${a_n}$ converges, to, say, c. Then there is a small interval (c-r,c+r) that contains all-but-finitely-many points of ${a_n}$. Then there are only finitely-many points in each of [a,c-r] and in [c+r,b]. Define the function d($a_n,b_n)=|a_n-b_n|$, in each of these subintervals and throw- out terms that are equal to each other. The function reaches a minimum m (since a finite set of points is compact, or simply because the min. of a finite set exists and is well-defined ) after having thrown out the terms between each other, the value of m is >0. It follows that there is a gap of width at least m between any two terms in the sequence, and this gap is not filled by any other terms in the sequence.
2)${a_n}$ diverges in [a,b]. Then ${a_n}$ is not Cauchy, so that there is some r>0 with
  $|a_n-a_m|$>r for all m,n integers. So there is a gap in [a,b] not covered by the sequence $a_n$
