Do universally connected spaces exist?

Question

Let's call a topological space $X$ connected iff it can not be represented as a disjoint union of two non-empty open sets.

Now, suppose $X$ and $Y$ are both topological spaces. Let's say $X$ is $Y$-connected iff $\forall x_0, x_1 \in X$ $\exists y_0, y_1 \in Y$ and a continuous function $f: Y \to X$ such that $f(y_0) = x_0$ and $f(y_1) = x_1$.

Those two variants of connectivity are connected in the following ways:

If $X$ is $Y$-connected and $Y$ is connected then $X$ is also connected.

Proof: Suppose, $X$ is not connected. Then it is a disjoint union of two non-empty open sets $U$ and $V$. Let's take $x_0 \in U$ and $x_1 \in V$. Then $Y$ will not be connected as a disjoint union of two non-empty open sets $f^{-1}(U)$ and $f^{-1}(V)$, Q.E.D.

Converse is however generally false - for example, $\{(t, \sin(\frac{1}{t})|t \in \mathbb{R})\} \cup \{(0, 0)\}$ is connected, but not $[0;1]$-connected.

If $Y$ is not connected, then all topological spaces are $Y$-connected.

Proof: Suppose, $Y$ is not connected. Then it is a disjoint union of two non-empty open sets $U$ and $V$. Now, suppose $X$ is an arbitrary topological space and $x_0, x_1 \in X$. Now we take $y_0 \in U$, $y_1 \in V$ and define $f$ as

$$f(y) = \begin{cases} x_0 & \quad y \in U \\ x_1 & \quad y \in V \end{cases}$$

Those $y_0, y_1$ and $f$ satisfy our conditions. That means, $X$ is $Y$-connected, Q.E.D.

If $X, Y$ and $Z$ are such topological spaces, that $X$ is $Y$-connected and $Y$ is $Z$-connected then $X$ is $Z$-connected.

Suppose $x_0, x_1 \in X$. Thes $\exists y_0, y_1 \in Y$ and and a continuous function $f: Y \to X$ such that $f(y_0) = x_0$ and $f(y_1) = x_1$. There also $\exists z_0, z_1 \in Z$ and a continuous function $g: Z \to Y$ such that $f(z_0) = y_0$ and $f(z_1) = y_1$. Then $f(g(z_0)) = x_0$ and $f(g(z_1)) = x_1$, which means $X$ is $Z$-connected, Q.E.D.

Now let's call a topological space $Y$ universally connected iff a topological space being $Y$-connected is equivalent to it being connected (equivalently, $Y$ is connected and all connected spaces are $Y$-connected)

My question is:

Do universally connected spaces exist?

Answer 1

No such space exists.

Assume $Y$ is universally connected. Now consider $\lambda$ to be any ordinal with the corresponding cardinality greater then that of $Y$. Let $X=\lambda\times[0,1)\sqcup\{\lambda\}$ with the lexicographical order/topology. The extended extremely long line. ;) The space $X$ is connected and in fact connected subsets of $X$ are intervals. In particular if $A\subseteq X$ is a connected subset containing the beginning $x_0$ and the end $x_1$ then $A=X$. This means that if $f:Y\to X$ is any continuous map with $\{x_0,x_1\}$ in the image then $f(Y)=X$. But this is impossible due to cardinality.