Why do you use triangle inequality in this proof? I just read this question (The non-empty intersection of two open discs contains an open disc.) and I don't understand how the triangle inequality lines complete the proof. Please be nice - I've just started learning topology but hopefully I'll get there!
Thanks in advance for your help.
 A: Say we have three points $x, y$, and $z$ in a metric space $X$ (a metric space is simply a set where we have a notion of a distance - i.e., I can tell you the distance between any two points in the space $X$). Say I know the distance $d(x,y)$ between points $x$ and $y$, and I know the distance $d(x,z)$ between $x$ and $z$. The question is, can I say anything about the distance between $y$ and $z$? It might seem like the answer is no, at first, but the triangle inequality says something very important: It tells us the largest possible value the distance $d(y,z)$ between $y$ and $z$ can be. More specifically, the triangle inequality tells us that $$d(y,z) \leq d(y,x) + d(x,z)$$
The triangle inequality gives us information about $d(y,z)$ that we wouldn't otherwise have. But now, how does this concept relate back to the problem at hand? Say that I have a ball $B_r(x)$ of radius $r>0$ centered at a point $x$. That is, $$B_r(x) = \{m \in X: d(m,x) < r\}$$. Similarly, say I have a ball $B_{r'}(y)$ of radius $r' > 0$ centered at a point $y \in X$. Say $z$ is a point in both balls. I want to show that there is a ball centered at $z$ that is contained in both of the previously mentioned balls (which are centered at $x$ and $y$).
Since $z \in B_r(x)$, we know that $d(z,x) < r$. But because $z \in B_{r'}(y)$, we also know $d(z,y) < r'$. Can we find a ball centered at the point $z$ contained in the intersection $B_{r}(x) \cap B_{r'}(y)$? My answer is yes! Consider the ball $B_{\epsilon}(z)$ centered at $z$ of radius $\epsilon = \min \{r - d(z,x),r' - d(z,y)\}$. Let's show that $B_{\epsilon}(z)$ is a subset of $B_r(x)$. Let $m \in B_{\epsilon}(z)$. I want to show that $m \in B_r(x)$. That is, I need to show $d(m,x) < r$. How am I going to do this?
Well, since $m \in B_{\epsilon}(z)$, I know that $d(m,z) < \epsilon$. I know something about the distance between points $m$ and $z$ (namely $d(m,z) < \epsilon$), and I also know something about the distance between points $z$ and $x$ (namely, $d(z,x) < r$). But, how in the world would I find something that tells us about the distance between $m$ and $x$?? Well, this is where the triangle inequality comes in!! The triangle inequality directly tells me that $$d(m,x) \leq d(m, z) + d(z, x)$$ Because I know $d(m,z) < \epsilon$, I can say that the right-hand side of the above inequality is less than $\epsilon + d(z,x)$. In other words, $$d(m,x) \leq d(m,z) + d(z,x) < \epsilon + d(z,x)$$ But by the way we defined $\epsilon$, I know that $\epsilon \leq r-d(z,x)$. Hence, I can say that $$d(m,x) < \epsilon + d(z,x) \leq (r- d(z,x)) + d(z,x) = r$$
Hence, $d(m,x) < r$ as we wanted to show (showing this  wouldn't have been possible without the triangle inequality!)
Showing that $B_{\epsilon}(z) \subset B_{r'}(y)$ is almost exactly the same. Hence, $B_{\epsilon}(z) \subset B_r(z) \cap B_{r'}(y)$
As a side note, I've been using the term "Balls" to refer to subsets of a metric space $X$. If you're working in the metric space $\mathbb{R}^2$, as the question you linked certainly is, a ball is also called a disc.
