Generally, showing something is not connected is easy in principle, you 'just' have to divide it into 2 non-empty open sets. Of course, finding those sets is the hard part, as usual having seen examples is the way to start and build an intuition for that.
Showing connectedness by the definition is hard, because it is often hard to prove that something with that many possibilities does not exist. Knowing that path-connectedness implies connectedness is a big help here, because showing that 2 arbitrary points can be connected by a path inside the set is often managable. The really hard cases are when something is not path conneced by still connected.
Problem 1) can be answered by first observing that each point $(q,y) \in \mathbb Q \times [0,1]$ is path connected to the point $(q,1)$, via
$$f: [0,1] \to C \text{ with } f(t)=(q,y+t(1-y)),$$
and two points on $\mathbb R \times \{1\}$ are obvioiusly path-connected to each other. Path connectedness is transitive, so any point in $C$ is path connected to any other via at most 2 points on $\mathbb R \times \{1\}$. So the $C$ in problem 1 is path connected and hence connected.
For problem 2) the argument is that a point with rational $x$-coordinate $(q,y)$ (via essentially the same kind of path function as in problem 1) is path connected to $(q,0)$ which is path connected to $(0,0)$. The same goes for a point with rational $y$-coordinate: $(x,q)$ is path connected to $(0,q)$ which is path connected to $(0,0)$. Again by transitivity of path connectedness, this makes the whole $C$ of problem 2 path connected and thus connected.
Problem 3) is IMO the hardest. Removel of $(1,1)$ from the $C$ of problem 1 makes the line $\mathbb R \times \{1\}$ become two rays instead: $(-\infty,1) \times \{1\}$ and $(1,\infty) \times \{1\}$, which are together not path connected.
Now it becomes important to recall the proof that path-connectedness implies connectedness. If a space is not connected, it can be partitioned into 2 open non-empty sets $A$ and $B$. If a path-connected space was not connected, one could find points $a \in A$ and $b \in B$, and then construct a path between them. It then turns out that this is not really possible.
Now assume that the $C$ from problem 3 is not connected, so it can be partitioned into 2 non-empty sets $A$ and $B$. If we can find path-connected 'commponents' in $C$, that means sets of points that are all path connected between themselves, the above proves that each such component must belong completely to either $A$ or $B$, you cannot have some points belonging to $A$ and others to $B$!
Using the same arguments as in problem one, one can now find that $C$ consists of 3 such path connected components:
$$C_1: \{(q,y) \in \mathbb Q \times [0,1]: q > 1\}\, \cup \, (1,\infty) \times \{1\}$$
$$C_2: \{(q,y) \in \mathbb Q \times [0,1]: q < 1\}\, \cup \, (-\infty,1) \times \{1\}$$
$$C_3: \{1\} \times [0,1)$$
We now have to consider only how these 3 components can be distributed among $A$ and $B$. Since neither $A$ and $B$ can't be empty, it means 2 components must go into one set, and one component is the sole other set.
Now it turns out that any ball around any point in $C_3$ will by necessity include points from $C_1$ (in the $\{(q,y) \in \mathbb Q \times [0,1]: q > 1\}$ part) and points from $C_2$ (in $\{(q,y) \in \mathbb Q \times [0,1]: q < 1\}$). That means $C_3$ must go into the same set as $C_1$ and $C_2$, which leaves the other set empty, which is forbidden.
So we finally see that although the $C$ of problem 3 is not path connected, but nonetheless connected.
Best Answer
At most countable may as well mean one -- this is the case for the open intervals. Then there is no contradiction with the conectedness.