For (a,b,c) that's more or less correct. In general, there doesn't have to be a voltage/current just because there is a short/open, there just can't be any voltage in a perfect short and there can't be any current in a perfect open.
Another way to re-word these two terms is that a short circuit has 0 resistance (R=0), and an open circuit has infinite resistance (R=infinity).
So in Ohm's law, \$V = IR\$.
If \$R = 0\$, then \$V = 0\$.
If \$R = \infty\$, then using some mathematical trickery:
$$
I = \lim_{R\rightarrow \infty} \frac{V}{R} = 0
$$
As far as the force analogy goes, if it's useful think about you pushing on a building. Just because you are applying a force doesn't mean the building is going anywhere. These type of analogies tend to break down when dealing with theoretical 0's and infinities, so I wouldn't rely too heavily on them but rather look at the mathematics.
KVL can be more generally stated as: the electric field is a conservative vector field. This can be expressed by the path integral:
\begin{align}
\int_{p1}^{p2} \vec{E} \cdot d\vec{l} = V_{p2} - V_{p1}
\end{align}
I.E. the integral is "path independent", and the result is the same no matter what path you take (so long as you start at p1 and end at p2). This can be 0, but it doesn't have to be. If \$p1=p2\$ (taking a closed path),
\begin{align}
\oint \vec{E} \cdot d\vec{l} = 0
\end{align}
This is why KVL even applies: KVL states that in any closed loop, the voltage across it is 0. It doesn't mean that the voltage must be 0 between any two points on that loop. It doesn't matter if there is any conductive path between two points. Having an ideal wire is just an extra statement that two distinct points are at the same voltage.
Best Answer
Converting current source into voltage source
You could read this article. If the current source confuses you, you could convert it into an equivalent voltage source using the following theory:
http://en.wikipedia.org/wiki/Source_transformation
Then,
The 2K ohm and 4K ohm resistors have zero current flowing through them (open circuit), and hence can be discarded in our voltage calculations. Since we need to find Vo, which is a potential difference, we can assume a reference node (top one) to be at 0V. Thus, the current division would be in the inverse ratio of the resistances in the parallel branches.
Hence, I1 = 15/(15+20) * 17 mA = 7.2857 mA.
Similarly, I2 = 20/(20+15) * 17 mA = 9.7142 mA.
Thus, Vo = I1 * 10 - I2*3 = 72.857 - 29.142 = 43.714 V.
This is your answer, and if you have any doubts, you could go through any of these articles:
http://en.wikipedia.org/wiki/Current_divider