Open and Closed balls are connected Prove that open and closed balls in the standard metric on $\mathbb{R}^2$
are connected.
I have seen a lot of stuff about being path connected, but I haven't learned that yet. I know $\mathbb{R}^2$ is connected. Is there a way to take that info and then conclude open balls are connected?
If I can show open balls are connected, than closed balls are just the closure of open balls (right? how do I show this?), so by a theorem in my book, the closure of a connected set is connected, so closed balls would be connected.
 A: 
I know $\mathbb{R}^2$ is connected. Is there a way to take that info and then conclude open balls are connected?

Yes: we can note that any open ball is homeomorphic to $\Bbb R^2$.  Note for instance that the map from the unit open ball centered at zero and $\Bbb R^2$ given in polar coordinates by
$$
(r,\theta) \mapsto \left(\frac{1}{1-r}, \theta\right)
$$
is a homeomorphism of the two spaces.

If I can show open balls are connected, than closed balls are just the closure of open balls (right? how do I show this?)

Yes: a closed ball is indeed the closure of the corresponding open ball.  To show this, it suffices to note that closed balls are closed and that every element of a closed ball is the limit of some sequence in the open ball.
A: The most obvious proof to me comes to me in terms of path connectedness, but let's try and avoid using the exact word but still use the idea. Let $B$ be an open ball and $\{U,V\}$ be a separation of $B$. Take $u_0\in U$ and $v_0\in V$. Consider the line $L$ formed by $u_0,v_0$ as end points. It should be easy to show that this line is contained in the ball (think of convexity; you have the explicit norm at your disposal here) and $L$ is connected. But, $U$ and $V$ must also form a separation of $L$ (take the intersections); a contradiction. An alternative way is to show that it is path connected because for any such choice of $u_0$ and $v_0$, the lines joining them to the centre of the ball are also contained in the ball and thus concatenating them, you get that the ball is path connected from which this follows trivially. 
The most obvious way to use the connectedness of $\mathbb R^2$ to me is to use homeomorphisms, and if path connectedness is something you haven't done I assume the same for this. 
To show that the closure of the open ball is the closed ball, use the two directions of showing one is a subset of the other. One containment is easy; for the other side, show that every point in the boundary of the closed ball is the limit of some sequence of points entirely contained in the open ball. 
