One way to state the Baire Category Theorem is as follows: If $X \neq \emptyset$ is a complete metric space, then $X$ is nonmeager as a subset of itself.
Then it was mentioned that the converse, that there are nonmeager spaces which are not complete, of this theorem is false. I would like to ask if the following reasoning is correct:
Proof: Take $(0,1) \subset \mathbb{R}$ and associate the usual metric $d(x,y) = |x-y|$ so that $((0,1),d)$ is a metric space. We claim this space is not complete. Take $\left(\frac{1}{n}\right)_{n=2}^\infty$. Then $\frac{1}{n} \rightarrow 0$ as $n \rightarrow \infty$, however $0 \notin (0,1)$. Thus, not complete. However, $$(0,1) = \bigcup_{n \geq 1}\left(0,\frac{1}{n}\right)$$ is nonmeager since $\left(0,\frac{1}{2}\right) \subset (0,1)$ clearly is not rare.
Edit: 
That counterexample is correct, but the argument is not. You claim that$$(0,1)=\bigcup_{n=1}^\infty\left(0,\frac1n\right),\tag1$$So what? Asserting that $(0,1)$ is a counterexample means that $(0,1)$ cannot be written as a countable union of rare subsets of itself, and the equality $(1)$ doesn't allow you to deduce that.
The space $(0,1)$ cannot be written as a countable union of rare subsets of itself because it is locally compact, and Baire's theorem holds if “complete metric space” gets replaced by “locally compact topological space”.