Why can a closed, bounded interval be uncountable? From what I have read, all finite sets are countable but not all countable sets are finite. As I understand it, 


*

*Countably Finite --- a one to one map onto $\Bbb{N}$ with a limited number of members 

*Countably Infinite --- a one to one map onto $\Bbb{N}$ with an unlimited number of members but that you can count in principle if given an infinite amount of time 

*Uncountably Infinite --- there is no one to one mapping onto $\Bbb{N}$. Even if you count, you will miss some of the members. And it is infinite. 


From this I gather that countable is not the same as finite. Countable is the one to one property with $\Bbb{N}$. Finite just means a limited number of elements.  
Now consider $[0,1]$ which is closed and bounded. 


*

*Bounded --- $\forall k \in [0,1]$ we have $k \leq 1$. Similarly all $k\geq0$. 

*Closed --- it contains the endpoints $0$ and $1$ 


Yet I read $[0,1]$ is uncountably infinite. So clearly, neither closure nor boundedness implies finiteness or countability.  
Question: 
Why can a closed, bounded interval be uncountable? 
It just seems like something that is bounded would be "more finite" than something that isn't. 
 A: I suspect you're conflating two meanings of "finite". Some sets are finite, meaning they have only finitely many elements.  An interval like $[0,1]$ is not such a set.  On the other hand, $[0,1]$ has finite length, which is a quite different matter.  As the other answers have explained, finite length does not imply finiteness (or even countability) in terms of the number of elements.
A: Certainly the interval $[0,1]$ is uncountably infinite; it has the same cardinality as $(0,1)$ and $\mathbb{R}$ and $\mathcal{P}(\mathbb{N})$ and many other sets which are commonly used in mathematics.
However, closed and bounded intervals such as $[0,1]$ do have a nice finiteness property called compactness: that is, if $\{ U_i \mid i \in I \}$ is any collection of open intervals such that $\bigcup_{i \in I} U_i = [0,1]$, then there is a finite subset $J \subseteq I$ such that $\bigcup_{i \in J} U_j = [0,1]$.
This is not true of $(0,1)$, since for instance if we define $U_i = (\frac{1}{i}, 1-\frac{1}{i})$ then $\bigcup_{i \in I} U_i = (0,1)$ but no finite subset $J \subseteq I$ has $\bigcup_{i \in J} U_i = (0,1)$.
A: Well, in some sense you are right. If a set is bounded, that means it has finite volume. If it is not bounded, it means it cannot be contained in any ball of finite volume (does not mean it has infinite volume though, e.g. consider the set of all $(k,0)$ for integer $k$ in $\mathbb{R}^2$).
Note that both finite and countable sets have 0 volume in real space...
A: In the sense of cardinality, $[0,1]$ and $\mathbb R$ are the same (ie they have the same cardinality) because you can find a bijection from $[0,1]$ to $\mathbb R$.
As you've noted, this has nothing to do with a set being closed or bounded
A: Let $[a,b]\subset\mathbb R$ be a closed interval-segment. What would happen if it were countable? 
(source: https://web.math.pmf.unizg.hr/~guljas/skripte/MATANALuR.pdf)
Then there would exist a bijection $f:\mathbb N\to[a,b]$. For the sake of understanding, let's treat it like a sequence, just like others have already suggested, and look it as nothing but points. I'll state we want to come to a contradiction according to the axiom of completeness and Cantor's axiom. We're dealing with nested-intervals.
Range/image of $f$: $$\mathcal R_f=[a,b]=\{f(n):n\in\mathbb N\}$$
Next step: $a=a_1,b=b_1$
$$\frac{a+b}{2}=\frac{a_1+b_1}{2}=\frac{2a_1+2b_2}{4}$$
Two cases:
$(i)$ $f(1)\leq\frac{a_1+b_1}{2}$ & $(ii)$ $f(1)\geq\frac{a_1+b_1}{2}$
Technical computations:
$a_1,b_1\in\mathbb R,a_1<b_1\implies$
$$\frac{a_1+b_1}{2}=\frac{a_1+a_1+2b_1}{4}<\frac{a_1+b_1+2b_1}{4}=\frac{a_1+3b_1}{4}\;\;\;\;\;(1)$$
$$\frac{a_1+b_1}{2}=\frac{2a_1+b_1+b_1}{4}>\frac{2a_1+a_1+b_1}{4}=\frac{3a_1+b_1}{4}\;\;\;\;\;(2)$$
Now we manipulate with this arbitrary function so as to get nonsense:
$(i)\;\&\;(1),\;\;a_2=\frac{a_1+3b_1}{4},\;b_2=b_1$ $$f(1)\leq\frac{a_1+b_1}{2}\implies f(1)\notin[a_2,b_2]$$
$(ii)\;\&\;(2),\;\;a_2=a_1,\;b_2=\frac{3a_1+b_1}{4}$ $$f(1)\geq\frac{a_1+b_1}{2}\implies f(1)\notin[a_2,b_2]$$
We managed to construct $[a,b]$ s.t. $f(1)\notin[a,b]$ neither in $(i)$ or $(ii)$.
We iterate this for $n\in\mathbb N$ and $$a_{n+1}=a_n\underline{\lor} b_{n+1}=b_n$$ and$$[a_{n+1},b_{n+1}]\subset[a_n,b_n],f(n)\notin [a_{n+1},b_{n+1}]\;\forall n\in\mathbb N$$
But$$\text{axiom of completeness}\implies\;\;\exists x\in[a,b]\cap[a_n,b_n]=[a_n,b_n]\;\forall n\in\mathbb N$$
and
$$f:\mathbb N\to[a,b]\;\text{is bijective}\implies\;\exists m\in\mathbb N\;s.t.\;f(m)=x$$
But (again) $$f(m)\notin[a_{m+1},b_{m+1}]\Rightarrow\Leftarrow$$
