Prove that no uncountable family of subsets of $\mathbb{N}$ is well-ordered by relation of inclusion.
I had two ideas how to do it. First was to show that if such family is uncountable, then it is not true that all proper beginning segment (is it correct english name for that?) is of the form $O(x)=\{y:y<x\}$ for some $x$.
Another idea was to assume that $(A,\subseteq)$ is such well-order. There are $a,b\in A, a\subset b$, such that between $a$ and $b$ there are uncountably many elements. Let's take subset $B$ which contains $a,b$ and all elements between them. Let's define sequence $z_n$, where $z_0=b$ and $z_{n+1}$ is element of set $\{x\in B:x<z_n$ and there are uncountably many elements between $a$ and $x\}$. By axiom of choice such sequence exists. Let's define $C=\{z_n:n\in\mathbb N\}$. Of course $C\subset A$ and $C$ does not contain minimal element. Therefore $(A,\subseteq)$ is not well-order.
Is it correct?
The second idea is on the right track. But you don't need contradiction at all. Instead prove the contrapositive. If $(A,\subseteq)$ is well-ordered, then $A$ is countable.
First, without loss of generality, $A$ does not have a maximal element (otherwise, remove finitely many sets from the top).
Let $A_\alpha$ be the $\alpha$th member of $A$. And define an injection from $A$ into $\Bbb N$. Since every point of $A$ has a successor, we can pick $n_\alpha$ as the least $n\in A_{\alpha+1}\setminus A_\alpha$.
Finally, prove that $\alpha\mapsto n_\alpha$ is an injection.