Let me observe that Noah's excellent answer generalizes to solve the full case.

**Theorem.** Every topological space is a continuous image of a dense linear order.

Proof. Let $\kappa$ be any ordinal number. Let $P$ be the linear order $\mathbb{Q}\times\kappa$, under the lexical order, which is obtained by replacing each ordinal less than $\kappa$ with a copy of the rational order $\mathbb{Q}$. This is a dense linear order, resembling the well-known *long line*, but with rationals and with $\kappa$ instead of the unit interval and $\omega_1$. You could call it the $\kappa$-long rational line. For each $\alpha\lt\kappa$, let $Q_\alpha=\mathbb{Q}\times\{\alpha\}$ be the copy of $\mathbb{Q}$ surrounding $\alpha$ in $P$. This is an open set, and they partition $P$ just as in Noah's argument.

Now, following Noah, let $X$ be a discrete space with $\kappa$ many points, and let $f:P\to X$ map $Q_\alpha$ to the $\alpha$-th point of $X$. This map is continuous, since the pre-image of every point is open. Thus, we have mapped $P$ continuously to the discrete space of size $\kappa$, and by composition, we can now map $P$ to any space of size $\kappa$. QED

The argument shows that every infinite topological space is a continuous image of a dense linear order of the same size.

Note that this proof uses the Axiom of Choice. But full AC doesn't seem required, since we didn't really need that $X$ was well-orderable, but rather only that it was linearly orderable. That is, if $X$ is a linearly orderable set, then by replacing each point with a copy of the rationals, we get a dense linear order, with a locally constant map back to $X$. So the discrete topology on $X$ is the continuous image of a dense linear order, and therefore any topology on $X$ is the continuous image of a dense linear order. (The assertion that every set admits a linear order is a weak choice principle.)