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).