I've found on wikipedia a proof but I don't really understand ot. If a topological space $S$ (order topology) is linear continuum it satisfies the next:
a) $S$ has the least-upper-bound property
b)For each $x \in S$ and each $y \in S$ with $x < y$, there exists $z \in S$ such that $x < z < y$
Property b) is trivial. To check property a), we define a map, $\pi_1 : I \times I \to I$ by $\pi_1 (x, y) = x$
This map is known as the projection map. The projection map is continuous (with respect to the product topology on $I \times I$) and is surjective. Let $A$ be a nonempty subset of $I \times I$ which is bounded above. Consider $\pi_1(A)$. Since $A$ is bounded above, $\pi_1(A)$ must also be bounded above. Since, $\pi_1(A)$ is a subset of $I$, it must have a least upper bound (since I has the least upper bound property). Therefore, we may let $b$ be the least upper bound of $\pi_1(A)$. If $b$ belongs to $\pi_1(A)$, then $\{b\} \times I$ will intersect $A$ at say $(b,c)$ for some $c \in I$. Notice that since $b \times I$ has the same order type of $I$, the set $(\{b\} \times I) \cap A$ will indeed have a least upper bound $(b,c')$, which is the desired least upper bound for $A$.
If $b$ does not belong to $\pi_1(A)$, then $(b,0)$ is the least upper bound of $A$, for if $d < b$, and $(d,e)$ is an upper bound of $A$, then $d$ would be a smaller upper bound of $\pi_1(A)$ than $b$, contradicting the unique property of $b$.
¿Can someone tell me why $b$ needs to belong to $\pi_1(A)$?
Actually, $b$ need not belong to $\pi_1(A)$ in general. For example, take $$A = [0, 1/2) \times [0, 1].$$ Then $\pi_1(A) = [0, 1/2)$, and $b = 1/2 \notin \pi_1(A)$. The proof doesn't claim this, and instead splits into two cases: if $b \in \pi_1(A)$ and if $b \notin \pi_1(A)$.
In the case where $b \notin \pi_1(A)$, they nominate the element $b \times 0$, and prove it is the supremum of $A$ by contradiction. In the above case, we have $1/2 \times 0$ as the supremum of $A$.