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 looses 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.