Every infinite set has an infinite countable subset? As the title says, that's all my question. Let me state it again:


  Is it true that every infinite set has an infinite countable subset?


It seems so trivial, my thought goes like this: pick an arbitrary element and denote it as $x_1$; pick the next one and denote it $x_2$, and so on.

Is my proof correct? Since it seems so simple, I'm not sure of it. 



To avoid any further confusion, the definitions used are:

Finite: In bijection with $\{1\ldots n\}$ for some $n$.
Infinite: Not finite.
Countably infinite: In bijection with $\Bbb N$.
Countable: Finite or countably infinite.
You are on the right track, however you cannot use the phrase "the next one".  The idea is:Let $S$ be an infinte set. Pick an element $x_1\in S$, then since $\{x_1\}$ is finite,
$S\setminus\{x_1\}\ne\emptyset$, so pick an element $x_2\in S\setminus\{x_1\}$ and so on.

Note the Axiom of Choice is involved here.
If you are using Dedekind's definition of infinity, then there must exist an injective, but not surjective function $f: S \to S$, where $S$ is your infinite set. Then there must also exist a countable $n\subset S$ that is order-isomorphic to $\mathbb {N}$ with the function $f$ serving as the successor function and any element outside the range of $f$ serving as the first element $0$ (or $1$). 

For details and formal proofs, see my posting "Infinity: The Story So Far" at my math blog.