Let $(L, <)$ be an uncountable Linear order then is the following claim true? For Every Uncountable Linear order $(L,<)$ there exist a uncountable subset $X$ of $L$ such that $X$ is well ordered by $<$
I need to prove/disprove above claim
Here is my attempt on this question :
Let $a \in L$, then we consider the initial segment $Y = \text{pred}(X,<,a)$ if this segment is countable then I can consider
$Z =L-Y$ this will be a well ordered set, as it will have $a$ as its least- member and will be uncountable
the problem is I am not sure how to produce such as a initial segment ,so I cannot proceed further in this question.
Since , I am a beginner in set theory it will be helpful if someone can give a proper solution to understand the subject better and I can use the techniques in solving future problems
 A: Hagen von Eitzen's example works, but isn't using the full power which well-orderings provide us:

Suppose $\alpha$ is a "co-well-ordering," that is, a linear order in which every nonempty subset has a maximum element. Then every ascending sequence in $\alpha$ is finite.

This is the same exact argument as the proof that every descending sequence in a well-ordering is finite. From this we then get:

Suppose $\alpha$ is a co-well-ordering. Then $\alpha$ has no infinite (let alone uncountable) suborders which are well-orders.

This is because if $\beta$ is an infinite well-ordering, then $\beta$ contains an infinite ascending sequence (think about the "first $\omega$-many elements").
So we can improve Hagen's example as follows:

If $\alpha$ is a co-well-ordering (not necessarily minimal) then $\alpha$ has no infinite (let alone uncountable) well-ordered suborder.

A: Let $L$ be the smallest uncountable ordinal, but $<$ is the reverse order.
Assume $X\subseteq L$ has  a minimal element $x_0=\min X$ (which is certainly the case if $X$ is well-ordered by $<$).
By construction, there are only countable may $y\in L$ with $y>x$, hence $X$ is countable.
