So I tried to prove the following statement:
If $(X, \leq$) and $(X, \geq)$ are well orders, then X is finite.
But I'm not sure wether it's entirely correct.
Proof:
A well-ordered set has the property that for every infinite decreasing sequence
$x_0 \geq x_1 \geq \cdots$ there is an $n$ such that for every $m > n$: $x_n = x_m$.
Since the set has an element $x_0$ so that for every $x \epsilon X: x \leq x_0$, we can chose the infinite decreasing sequence
$x_0 \geq x_1 \geq \cdots$ which contains every element of $X$ and will stop because $(X, \leq)$ is a w.o.. Say this 'least element' is $x_N$, then there is a bijection $\{0,..,N\} \rightarrow X$, so $X$ is finite.
It seems correct to me, but I'm not sure if I even understand the statement that I have to prove. Isn't $[0,1]$ as subset of $\mathcal{R}$ a w.o. that's infinite?
To cut down on the number of unanswered questions, here we go.
You certainly had a good idea! The only issue is with your rigor. Let's explicitly define the sequence you're using.
Since $X$ is non-strictly well-ordered by $\ge,$ then either $X$ is empty (in which case, we're done) or there is an element $x_0\in X$ such that $x_0\ge x$ for all $x\in X.$ From there, we proceed by recursion. If we've defined $x_n$ for some $n\in\omega,$ then we consider the set $X\setminus\{x_k:0\le k\le n\}.$ If this set is empty, we define $x_{n+1}:=x_n.$ Otherwise, since $X$ is non-strictly well-ordered by $\ge,$ then there is an element $x_{n+1}\in X\setminus\{x_k:0\le k\le n\}$ such that $x_{n+1}\ge x$ for all $x\in X\setminus\{x_k:0\le k\le n\}.$
In this way, we have defined a sequence of elements of $X$ such that $x_0\ge x_1\ge x_2\ge\dots.$ Since $X$ is non-strictly well-ordered by $\le,$ then there exists an $n$ such that $x_m=s_n$ for all $m>n.$ Let $n_0$ be the least such $n.$ Then the function $\{n\in\omega:n\le n_0\}\to X$ given by $n\mapsto x_n$ is a bijection, and so $X$ is finite.