Can I prove the Baire Category Theorem in this bizarre way? I'm trying to prove the Baire Category Theorem, which states that in a complete metric space $X$, the countable union of nowhere dense sets has empty interior.
In a metric space, $A$ being nowhere dense (having empty interior) implies that $\sup\limits_{p\in \overline{A}} d(p,(\overline{A})^c)=0$, where $A^c$ is the complement of $A$ in $X$. However, more weakly, it implies that $\sup\limits_{p\in A} d(p,A^c)=0$. Let us call this second function $f(A)$. 
Let us now number the countable set of nowhere dense sets $\{A_1,A_2,\dots,\}$. Let $B_i=A_1\cup\dots \cup A_i$. Now consider the sequence $\{B_1,B_2,\dots\}$. Clearly, $f(B_i)=0$ for all $i\in\Bbb{N}$. 
Can we say that $\lim\limits_{i\to\infty} B_i=\cup_{i} A_i$? Also, can we say that $f(A)$ is a continuous function in the weird topological space that contains subsets of $X$, and hence we can pass on to the limit and conclude that $f(\cup_i A_i)=0$?
How does the completeness of $X$ play into this? Moreover, how does the fact that $\sup\limits_{p\in \overline{A_i}} d(p,(\overline{A_i})^c)=0$ play into this?
 A: 
Can we say that $\lim\limits_{i\to\infty} B_i=\cup_{i} A_i$? 

Sure, you can think of it that way if you like.

Also, can we say that $f(A)$ is a continuous function in the weird topological space that contains subsets of $X$, and hence we can pass on to the limit and conclude that $f(\cup_i A_i)=0$?

No, that's not true in general.  For instance let $X=[0,1] \cup \{50\}$ with the Euclidean metric. Let $A_1 = \{0\}$ and $A_n = (1/n, 1]$ for $n \ge 2$.  Then $f(A_1 \cup \dots \cup A_n) \le 1$ for every $n$, but $f(\bigcup_{n=1}^\infty A_n) = 50$.  
It is true that $f(A)=0$ if and only if $A$ has empty interior.  But this is weaker than being nowhere dense, and so this won't help you prove the Baire category theorem.  It is certainly possible that a countable union of sets with empty interior has nonempty interior: enumerate the rationals as $q_1, q_2, \dots$ and let $A_n = \mathbb{R} \setminus \{q_n, q_{n+1}, \dots\}$.

How does the completeness of $X$ play into this?

It doesn't, really.
