Is there a "deep line" topological space in analogue to the "long line" $\omega_1\times[0,1)$?

$\def\RR{\mathbb{R}}$Let $X = \RR^{\omega_1}$; the set of all maps $\omega_1 \to \RR$. Make $X$ into an abelian group by pointwise addition. Let $f\in X$ be nonzero. Let $a$ be the least element of $\omega_1$ for which $f(a) \neq 0$. (Since $\omega_1$ is well ordered, there is a least such $a$.) Define $f$ to be positive if $f(a) >0$ and negative if $f(a) < 0$. This makes $X$ into an ordered abelian group. Equip $X$ with the order topology. Since I built $X$ as the order topology on an ordered abelian group, $X$ is homogenous and one-dimensional in your sense.

Let $f_n$ map the minimal element of $\omega_1$ to $n$ and map everything else to $0$. Then the sequence $f_n$ marches all the way to the end of $X$, so this line is not "long".

I claim that $X$ is deep. Let $U_i$ be a countable collection of open sets in $X$ containing $0$; we must show that $\bigcap U_i$ is not $\{ 0 \}$. Replacing each $U_i$ by a smaller open set, we may assume that $U_i$ is of the form $(f_i, g_i)$ with $f_i < 0 < g_i$. Let $a_i$ be the least element of $\omega_1$ for which $f_i(a_i) \neq 0$, so $f_i(a_i) < 0$. Similarly, let $b_i$ be minimal with $g_i(b_i) > 0$. Countable subsets of $\omega_1$ have upper bounds so we may choose $c \in \omega_1$ with $a_i < c$ and $b_i < c$ for all $i$. Let $f(c) = 1$ and $f(d) = 0$ for all other $d \in \omega_1$. Then $(-f,f) \subset (f_i, g_i)$ for all $i$, so the intersection contains the whole interval $(-f,f)$.