It seems I'm going to continue a sequence of dumb set-theoretic questions (here's the first one I asked yeserday: Simple properties of cardinal numbers). Sorry again if it's a duplicate.
I'm interested if $n\leq\mathfrak{m}$ where $n$ is a finite and $\mathfrak{m}$ is an infinite cardinal, and can it be proved without $AC$? My attempts were the following. Let say we want to prove $2\leq\mathfrak{m}$. Consider a set $M$ with cardinality $\mathfrak{m}$. It follows (?) from the definition of infinite sets that: \begin{align} \exists(a\in M)\exists(b\in M):a\neq b \end{align} Thus there is a subset of $M$ with cardinality $2$. The natural way to generalize these thoughts is to use the induction. Thus after observing $0$-case we're supposing there is a subset $X$ of $M$ set with cardinality $n$. Then we conclude $X\neq M$, and finally: \begin{align} \exists(a\in M):\overline{\overline{\{a\}\cup X}}=n+1 \end{align} The problem is I'm not sure if this pseudo-proof is valid. This raises the question again...
The proof is fine. You're appealing to induction. You would only need to choice if you want to claim that from this proof follows that every infinite set has a countably infinite subset.
There are two key facts here:
That means that if we managed to somehow find a subset of size $n$, then there is a subset of size $n+1$. And by induction that means there are subsets of every finite size.