The definition of countable is that a set is countable precisely when there is a bijective map from it to a subset of the natural numbers. See for example wikipedia.
Therefore your question is answered immediately by definition.
The more interesting question, which is what I think you really mean to ask, is to show that any countable set is either finite, or there is a bijective correspondence between it and $\mathbb{N}$.
Suppose we have an injective map $f\colon S\to \mathbb{N}$. Suppose that $S$ is not finite. We will construct a bijection $g\colon \mathbb{N}\to S$.
Firstly we describe an inductive process for picking elements $s_0,s_1,s_2,\cdots\in S$.
We know $f(S)$ is not empty, as $S$ is not finite (hence not empty), and given $s\in S$ we have $f(s)\in f(S)$. By the well-ordering principle we may pick $x\in f(S)$ minimal in $f(S)$. As $f$ injective, there is a unique $s_0\in S$ such that $f(s_0)=x$.
Once $s_0,s_1,\cdots,s_{i-1}\in S$ have been selected, we select $s_i\in S$ as follows. The set: $$X_i=\{x\in f(S)| x>f(s_{i-1})\},$$ is non-empty, as otherwise the image of $f$ would be a finite set, so $S$ would be finite. By the well-ordering principle we may pick $x\in X_i$ minimal. As $f$ injective, there is a unique $s_i\in S$ with $f(s_i)=x$.
Now we claim the function $g\colon \mathbb{N}\to S$ mapping $i\mapsto s_i$ is bijective.
Note that by construction, for $i>0$ we have $f(s_i)\in X_i$ so $f(s_i)> f(s_{i-1})$. Then by induction on the size of $j-i$, we have if $j>i$ then $f(s_j)>f(s_i)$. Thus $g$ must be injective.
Now pick an arbitrary $s\in S$. To show that $g$ is surjective, we must show that $s=s_i$ for some $i\in \mathbb{N}$. We know $f(s_i)$ is an increasing sequence, so for some $j\in\mathbb{N}$, we have $f(s)<f(s_{j})$, which is minimal in $X_j$, so $f(s)\notin X_j$.
To complete our proof we will prove by induction that for $j>0$:$$f(S)=\{f(s_0),f(s_1),\cdots,f(s_{j-1})\}\cup X_j$$
Thus we have that $f(s)=f(s_i)$ for some $i<j$ and $s=s_i$.
We selected $f(s_0)$ minimal in $f(S)$, so for every element $x\in f(S)$, either $x=f(s_0)$ or $x>f(s_0)$ so $x\in X_1$:$$f(S)=\{f(s_{0})\}\cup X_1$$
For the inductive step, we need only note that we selected $f(s_{i-1})$ minimal in $X_{i-1}$, so any element of $X_{i-1}$ is either $f(s_{i-1})$, or greater than $f(s_{i-1})$, so lies in $X_i$:
$$X_{i-1}=\{f(s_{i-1})\}\cup X_i\qquad\qquad (1)$$
Thus for $i\geq 2$: \begin{eqnarray*}f(S)&=&\{f(s_0),f(s_1),\cdots,f(s_{i-2})\}\cup X_{i-1}\\&=&\{f(s_0),f(s_1),\cdots,f(s_{i-2})\}\cup \{f(s_{i-1})\}\cup X_i\end{eqnarray*}
Here the first equality comes from the inductive hypothesis, and the second equality comes from $(1)$.
Best Answer
There are different ways to think about the size of a set. In the case of the real numbers, and specifically intervals, we can talk about their length (and generally, their Lebesgue measure in the case of measurable sets).
If you think about the real numbers as a model of time or space, then the distance between you and the screen through which you are reading this is a finite interval. But in this model, based on the real numbers, it is an uncountable interval, not a finite set.
One thing to remember about terminology, is that it should highlight to the reader or listener something about a certain relevant property. In the case of intervals, we already know they all have the same cardinality (in the case of non-degenerate intervals). So we can use "finite" or "infinite" to talk about their length (and formally, their measure). Thus setting the importance on that aspect, rather than their cardinality.