On the proof that if a set is open and arc-connected then it's connected by broken lines. Let $(X,d)$ be a metric space, and $\gamma : [a,b] \rightarrow X$ be a continuous function then $\gamma([a,b])$ is called a continuous arc.
I want to prove that if a set $C$ is open and arc-connected then it's connected by broken lines.
$A \in X$ is said to be arc-connected if $\forall{x,y} \in A$ there exists a  $\gamma:[a,b] \rightarrow X$ s.t. 


*

*$\gamma$ is continous

*$\gamma(t) \in A \forall t \in [a,b]$

*$\gamma(\alpha) = x, \gamma(b) = y$.


My procedure would be pick any $x,y \in C$ s.t. they are arc connected then there exist an open ball $B(x,\epsilon) \subset A$ now $B(x,\epsilon) \cap \gamma(t) \ne \emptyset$ because $\gamma(t)$ is continous and $\gamma(\alpha) = x$.
Take any point $x_1$ of the intersection different from x and connect it by a broken line then there exist an open ball $B(x_1,\epsilon_1) \subset A$ s.t. $B(x_1,\epsilon_1) \cap \gamma(t) \ne \emptyset$ taking any point $x_2$ of the intersection and repeating this procedure eventually connects $x$ and $y$ by broken lines.
This does not seem too precise to me, how would you prove it?
PS on little (big) doubt I had in class: when drawing the $\gamma$ the professor would sometimes draw a curve that doubled back or intersected itself saying that it is ok because we are in a metric space, would it not cease to be a function? 
 A: Since the other answers already address your idea, I will try to focus on solving the problem.
We will abuse of connectedness. Define the equivalence relation:
$x \sim y \text{ if }  y \text{ and } x \text{ can be connected by a broken line}.$
This is obviously an equivalence relation. Now, since the set is open, every equivalence class is easily seen to be open. Therefore, if you suppose that there is more than one equivalence class (obviously, there is one), you get a separation of your space, contradicting connectedness.
A: Your idea is good but needs a bit of modifying. First of all, if you take open ball and take intersection of the arc with the balls boundary for the center of the next ball, who can guarantee that the radii won't converge to $0$ too fast so you will never be able to reach $\gamma(b)$? But, for any $x\in[a,b]$ you can choose an open ball $B_x = B(\gamma(x),r_x)\subseteq C$. This is obviously an open cover of $\gamma([a,b])$, which is compact as continuous image of compact set, thus has finite subcover $\{B_{x_1},B_{x_2},\ldots,B_{x_n}\}$. Now use convexity of each of the balls to connect $\gamma(a)$ to $\gamma(b)$ with piecewise linear path.
Also, on your question on self-intersections. This will happen whenever $\gamma\colon [a,b]\to X$ is not injective function. Do not confuse the image of an arc with graph of a function.
A: First of all a comment: I think what you mean by a "broken line" is a continuous, piecewise straight path. However, to define a straight line, you need to be in Euclidean space (or at least a Riemannian/Finsler manifold). So let's assume $(X,d)$ is $\mathbb{R}^n$ with the Euclidean metric.
Your idea looks fine, but you have to be careful with the statement that you repeat picking points $x_{i+1} \in B(x_i;\epsilon_i)$: how do you know that the sequence of points eventually reaches $y$?
Instead, you can use the fact that the image of $\gamma$ is compact: take all points $z \in \gamma([a,b])$ and balls $B(z;\epsilon_z) \subset C$, then choose a finite cover. Draw a picture in 2D to see how to now pick your points $x_i$ and connect these by straight lines.
Finally, on your professor's remark that $\gamma$ can double back upon itself: that is no problem since if $\gamma(t_1) = \gamma(t_2)$, it's only the values of $\gamma$ that are duplicate, not the domain points $t_1 \neq t_2$!
