Are all metric spaces topological spaces? I think this is true but i cannot prove it. Any answer or hints are welcome. 
I have tried to start with $\mathbb{R}$ with euclidean metric. We may consider $\tau :=\{\emptyset,\mathbb{R}\}$ and obvious it is a topology in $\mathbb{R}$. but in general, there are many different kinds of metric spaces so I am not sure how to prove it.
 A: Yes. In a metric space $(X,d)$, the metric defines the open sets as follows: A set $U$ is considered open iff for all $x \in U$, there exists some $\epsilon>0$ such that $B(x,\epsilon) \subset U$. ($B(x,\epsilon) = \{y \mid d(x,y) < \epsilon \}$). The topology induced by the metric consists of these open sets.
You need to prove that this is indeed a topology.
A: For every metric space $(X,d)$ there is a natural induced topological structure. This natural choice of topology $\tau$ is given by the topology generated by open balls. That is, $\{ B(x,r) : x \in X, r > 0\}$ forms a subbase for $\tau$.
The topology is called induced because the open sets determined by $d$ and the open sets determined by $\tau$ agree.
A: You can equip a space metric with various topologies. The most common is generated by the balls. Take the the lower topology containing the balls. Also you can take a trivial topology like you said in $\mathbb R$. Is the same.
A: As stated the answer is 'no'. A metric space is not a topological space. However, every metric space gives rise to a topological space in a rather natural way. This is the well known construction that takes a metric space $X$ and constructs the topology on $X$ where a set $U$ is open precisely when for every $x\in U$ there exists some $e>0$ such that the open ball $B_e(x)$ is contained in $U$. 
Several comments are due. First, this process loses information. For instance, there exists infinitely many metrics on $\mathbb R$ such that all of them produce the same topology of open balls. So, only knowing the induced topology does not allow you to recover the metric. 
The construction mentioned is most clearly understood in the context of the categories of metric spaces and topological spaces. Let $Met$ be the category of all metric spaces and continuous mappings and let $Top$ be the category of topological spaces. The construction above is the object part of a functor $Met\to Top$, it sends any function between metric spaces to itself considered as a function between the associated topological spaces. Now, this functor is trivially faithful but interestingly it is full. This last property says that a function $f:X\to Y$ between metric spaces is continuous via the usual $\epsilon-\delta $ definition if, and only if, the same function $f$ considered now to be between the associated topological spaces is continuous (in the sense that the inverse image of an open is open). 
This last remark shows why the topology of open balls is the one most commonly used. It establishes a strong relation between metric spaces and topological space. However, this resulting functor is not an equivalence of categories. It does not even have a left or a right adjoint. This failure of the functor $Met\to Top$ to have any sort of inverse is a way to measure (or see) how different metric spaces are from topological spaces. 
Later Addition: As it turns out, metric spaces and topological spaces are equivalent, if metric is interpreted broadly enough. The details can be found here, alg. univ. (to appear). 
