Stronger form of connectedness than path-connectedness Given a topological space $C$ and points $c_0, c1\in C$ we say that a topological space $(X,\tau)$ is $(C,c_0,c_1)$-connected if and only if for all $x,y\in X$ there is $f:C\to X$ continuous with $f(c_0) = x$ and $f(c_1)=y$.
(In this language, path-connectedness equals $([0,1],0,1)$-connectedness.)
What is an example of a space $C$ and points $c_0\neq c1\in C$ such that


*

*$(C,c_0,c_1)$-connectedness implies path-connectedness, and 

*for every infinite cardinal $\kappa$ there is a topology on $\tau$ on $\kappa$ such that $(\kappa,\tau)$ is path-connected, but not $(C,c_0,c_1)$-connected


?
 A: Let $C$ be the space nº74 ("double origin topology") in Steen & Seebach's Counterexamples of Topology, chosen because it is the only one listed there that is T2 and path-connected but not T3:

$C$ consists of the set of points of the plane $\mathbb{R}^2$ together with an additional point $0^*$.  Neighborhoods of points other than the origin $0$ and the point $0^*$ are the usual open sets of $\mathbb{R}^2\setminus\{0\}$; as a basis fo neighborhoods of $0$ and $0^*$, we take $V_n(0) = \{(x,y) : x^2+y^2 < 1/n^2 \land y>0\} \cup \{0\}$ and $V_n(0^*) = \{(x,y) : x^2+y^2 < 1/n^2 \land y<0\} \cup \{0^*\}$.

In other words we have replaced the origin in the plane by an "upper origin" $0$ and a "lower origin" $0^*$, the neighborhoods of the upper origin being sets containing an open half-disk centered at the origin, plus the upper origin itself, and similarly for the lower origin.
The space $C$ is Hausdorff, but not T2½ because $0$ and $0^*$ do not have disjoint closed neighborhoods; in particular, it is not T3 or T3½.  So there is no continuous function $C \to \mathbb{R}$ taking different values on $0$ and $0^*$.
Let $c_0 = 0$ and $c_1 = 0^*$.  Then there is a path connecting $c_0$ and $c_1$: indeed, there is a path connecting $c_0$ to, say, $(1,0)$, and one connecting $(1,0)$ to $c_1$ (we can even find "arcs", i.e., injective paths, if we want).  If we have a continuous function $C \to X$ taking $c_0$ to $x$ and $c_1$ to $y$, then right-composing it with the path just mentioned gives a path connecting $x$ and $y$: so $(C,c_0,c_1)$-connectedness implies path-connectedness.  On the other hand, $\mathbb{R}$ is not $(C,c_0,c_1)$-connected because of what was said in the previous paragraph.
I think this answers the question, with the additional constraint that $C$ is Hausdorff and $c_0 \neq c_1$.  (I just noticed that Simon Henry had the same idea in the comments.)
However, it doesn't answer the question that I think should have been asked, namely to also require $C$ itself to be $(C,c_0,c_1)$-connected (in the above example, it's pretty clear that it's not).
