Connectedness in the language of path-connectedness
No such space $C$ can exist.
We will derive a contradiction from the assumption that $C,c_0,c_1$ as desired exists.
Let κ be any cardinal greater than $|C|$. View $\kappa$ as an ordinal. For each β in κ add a copy of the unit interval between β and β+1, and add a point ∞ at the end. The resulting "Very Long Line" $L_\kappa$ is dense and complete, hence connected as a topological space.
As $L_\kappa$ is connected, there must be a continuous map $f$ from $C$ into $L_\kappa$ whose image $L'$ contains $0$ and $\infty$. Let $b\notin L'$. Then the map $h$ that sends everything below $b$ to $0$ and everything above $b$ to $1$ is continuous from $L'$ to the discrete space $\{0,1\}$.
So the map $h\circ f:C\to \{0,1\}$ witnesses that $\{0,1\}$ is connected, a contradiction.
(This is just a variant of Helene Sigloch's earlier argument.)
First, there can't be a path from $c_0$ to $c_1$, else a continuous map would give a path from $x$ to $y$. By the same argument, $C$ is not allowed to have finitely many or countably many path components. If one can build something like the topologist's sine curve on a very long line of arbitrarily large cardinality, the number of path components of $C$ has to be larger or equal than any cardinal number. There is no largest cardinal number. Hence, the set of points of $C$ is no set.
Thus: If there exist long lines in every cardinality, the answer to your question is no.
Edit: OK, if $x$ and $y$ are uncountably far apart, there is no path from $x$ to $y$. Call the set of points between $x$ and $y$ the "long path" from $x$ to $y$. It contains an arbitrarily large number of connected non-path-connected, pairwise disjoint sub"paths". Say the number is $\kappa '$. The connected components of their preimages under $f$ also have to be non-path-connected and pairwise disjoint. Hence, the universal space $C$ that we are looking for needs to have at least $\kappa '$ path components, hence at least $\kappa '$ points. If the "very long line" exists for every cardinality, the universal space $(C, c_0,c_1)$ can't exist.
Edit: Goldstern's argument is much simpler than this. Just for completeness: Very long lines exist (see Goldstern's comment) and so the answer is definitely No.
Edit: The concept of homotopy theory with respect to a big interval is actually worked with. I just stumbled over a preprint by Penrod in the field and the foundations seem to be laid by Cannon-Gonner [J. Cannon and G. Conner, The big fundamental group, big Hawaiian earrings, and the big free groups, Topology Appl. 106 (2000)].
I claim that for each cardinal $\lambda$, there is a connected space $C$ and $c_{0},c_{1}\in C$ such that whenever $|X|<\lambda$, then $X$ is connected if and only if for all $x,y\in X$ there is some continuous map $f:C\rightarrow X$ with $f(c_{0})=x$ and $f(c_{1})=y$.
Let $I$ be an index set and let $(X_{i},x_{i},y_{i})_{i\in I}$ be an enumeration of all the connected spaces with two points of cardinality less than $\lambda$ up to an isomorphism preserving the $x_{i}$ and $y_{i}$. Let $C=\prod_{i\in I}X_{i}$, let $c_{0}=(x_{i})_{i\in I}$, and let $c_{1}=(y_{i})_{i\in I}$. Then $C$ is connected. Therefore if $X$ is a space and for each $x,y\in X$ there is some $f:C\rightarrow X$ with $f(c_{0})=x,f(c_{1})=y$, then $X$ is connected. On the other hand, if $X$ is connected, $|X|<\lambda$ and $x,y\in X$, then $(X,x,y)\simeq(X_{i},x_{i},y_{i})$ for some $i\in I$. However, the projection onto the $i$-th coordinate $\pi_{i}:C\rightarrow X_{i}$ is a continuous mapping with $\pi_{i}(c_{0})=x_{i},\pi_{i}(c_{1})=y_{i}$.