What does it mean to induce a topology? I am reading the definition of "metrizability" which states that if there exists a metric $d$ on set $X$ that induces the topology of $X$, then it is metrizable. My question is how can we possibly know what topology is being induced by metric $d$ which is supposed to be a distance function. How can the distance function and collection of open sets be related?
 A: Suppose that $\langle X,d\rangle$ is a metric space. Let
$$\mathscr{B}=\{B(x,r):x\in X\text{ and }r>0\}\;,$$
where $B(x,r)=\{y\in X:d(x,y)<r\}$ is the open ball of radius $r$ centred at $x$. It is not hard to verify that $\mathscr{B}$ is a base for a topology $\tau$ on $X$. By definition the topology $\tau$ generated by the base $\mathscr{B}$ is
$$\tau=\left\{\bigcup\mathscr{U}:\mathscr{U}\subseteq\mathscr{B}\right\}\;,$$
the family of sets that are unions of members of $\mathscr{B}$. In this case that means that $\tau$ contains the sets that are arbitrary unions of open $d$-balls. These are the open sets of the topological space $\langle X,\tau\rangle$, and $\tau$ is the topology on $X$ inducted by the metric $d$.
A: As Nate pointed out, the definition of the relationship is that you take the topology generated by the open sets, i.e. finite closures and arbitrary unions over all sets of the form $\{x \in \mathbb{R}:\; |x| < \rho\}$. But I don't think that's what you're asking about, since you seem to understand that a metric is induced on a topology, just not why.
Topological intuition takes some time to get accustomed to. You might picture it as the structure which plays an "in-between" role. Think of the set $\mathbb{R}$ as just a set of elements. It has little structure. Now think of $\mathbb{R}$ as a metric space. It now has a geometry, paths of least distance between two points (geodesics -- in this case, straight lines), and a whole bunch of structure associated to the notion that $d(x,y) = |x-y|$. But in between lies structure that takes no notice of distance...
This structure is the topology of your set, $\mathbb{R}$. You know what open sets are, presumably. [If not, think intervals $(a,b)$, or countable disjoint unions of them -- those are all the open sets in $\mathbb{R}$, a non-trivial fact.] The topology doesn't understand the difference between $(-1,1)$ and $(-\infty,\infty)$, at least not beyond different labelling. The reason is that there is a homeomorphism from $(-1,1)$ to $(-\infty,\infty)$ (hint: there is also one from $(-\pi/2,\pi/2)$ to $(-\infty,\infty)$ -- pick a nice trig function).
A homeomorphism is a transformation which preserves the topology of the space. When you think of the topology on a space, think of "up to homeomorphism". It is continuous and has a continuous inverse, and is bijective. Exercise: show that a homeomorphism $f$ takes open sets to open sets.
So what does all this have to do with the metric? First, open sets give you a notion of "closeness." For instance, you can define convergence of a sequence of points: say $x_n$ converges to $x$ if for any open set $A$ containing $x$ there is some $N$ such that $x_n \in A$ for all $n \geq N$.
The idea of things getting smaller and smaller (or open sets $d(x,y) < \delta$) reflects "fine-grain" structure. Topology, first and foremost, gives you a basic "map" of the space. The finer your topology, the better you know where you are. It's like telling you the continent I live in versus telling you the city. 
It's important to note that specifying which open set you're in tells you about where you are, and where you can't be. For instance, if $A = (-\infty,0)$, then $x \notin A$ means that $x$ has to be on the "right" side of $0$, i.e. $x \geq 0$. That's a geometric property. If I then tell you $x \notin (1,\infty)$, you know more or less where $x$ is -- more or less how far you are from, say, $3/4$.

A more significant consequence of the above property is that it tells you how open sets are "glued together". The definition of a topology tells you which open sets are which and which give you, say, a non-empty intersection (and which don't!). So for instance, if I took the real line $\mathbb{R}$ and considered $\pm\infty$ to be an "open set" (two elements in one open set! that's possible! but you lose a property), then I would effectively be glueing the two ends of of the real line (think of a rope) together, forming a circle (actually I might be getting something technical wrong here, kind of brain-dead, don't quote me on this).
The relationship to the metric is therefore significant, because you know about being "close" to other points, but you're unable to really quantify how close that really is. That's geometry.
An excellent exercise is to prove that $f$ is continuous (in the $\varepsilon-\delta$ sense) if and only if $f^{-1}(U)$ is open for any open set $U$. Once again, continuity is a property of closeness and it should be related to the topology (just like sequences converging). It's truly remarkable how good the definition of a topology and an open set are.
edit: how can I scale images?
A: Let $d$ be a metric. Then the topology induced by $d$ is the topology where the open sets are precisely the sets $U$ with the property that for every $u \in U$ there exists an $\epsilon > 0$ such that $B_\epsilon(u) \subset U$, where $B_\epsilon(u)$ denotes the open ball with radius $\epsilon$ and centre $u$. 
A: What book are you reading? It must have been carelessly written, if the definition of "metrizable space" is not preceded by a definition of "topology induced by a metric". Here is the missing definition: A set $U$ is open in the topology induced by a metric $d$ if and only if, for each point $a$ in $U$, there is a real number $\varepsilon\gt0$ such that $\{x:d(x,a)\lt\varepsilon\}\subseteq U$. (In other words, just the same way you define open sets in $\mathbb R^n$ using the ordinary Euclidean metric.) The topology induced by $d$ is just the collection of all open sets, as defined above.
