Let me give you some extra tools to use, here. Some details are left to you to verify.
Definitions: A subset $A$ of a topological space $X$ is said to be nowhere-dense (in $X$) if for every non-empty open subset $U$ of $X$ there is a non-empty open subset $V$ of $X$ such that $V\subseteq U$ and $V\cap A=\emptyset$. (You should be able to prove that this is equivalent to your definition of nowhere-dense.) A topological space $X$ is said to be a Baire space if every countable union of nowhere-dense subsets of $X$ has empty interior.
Lemma 1: $X$ is a Baire space if and only if every countable intersection of open dense subsets of $X$ is dense in $X$.
Proof: Suppose $X$ is a Baire space and $\{G_n\}_{n=1}^\infty$ is a countable collection of open dense sets, so each $X\setminus G_n$ is nowhere dense. (Why?) Thus, for any non-empty open set $U$, we have that $$U\nsubseteq\bigcup_{n=1}^\infty (X\setminus G_n)=X\setminus\left(\bigcap_{n=1}^\infty G_n\right),$$ so there must be some point of $\bigcap_{n=1}^\infty G_n$ in $U$. Thus, $\bigcap_{n=1}^\infty G_n$ is dense.
Suppose countable unions of open dense sets are dense, and let $A_n$ be nowhere dense for each $n$, so each $X\setminus\overline{A_n}$ is open and dense. (Why?) Take any non-empty open $U$, so that $U$ intersects $$\bigcap_{n=1}^\infty\left(X\setminus\overline{A_n}\right)=X\setminus\left(\bigcup_{n=1}^\infty\overline{A_n}\right)\subseteq X\setminus\left(\bigcup_{n=1}^\infty A_n\right),$$ so cannot be a subset of $\bigcup_{n=1}^\infty A_n$. Thus, $X$ is a Baire space. $\Box$
Lemma 2: $X$ is a Baire space if and only if every countable union of closed nowhere-dense subsets of $X$ has empty interior. (I leave the proof to you. Lemma 1 and DeMorgan's Laws should help.)
Lemma 3: If $X$ is a non-empty complete metric space with metric $d:X\times X\to\Bbb R$, and $F$ is a non-empty closed subset of $X$, then $F$ is a complete metric space with metric $d_F$ defined by $d_F(x,y)=d(x,y)$ for all $x,y\in F$. (I leave the proof of this to you.)
Proposition: Assuming BCT holds, every non-empty complete metric space is a Baire space.
Proof: Suppose by way of contradiction that $X$ is a non-empty complete metric space, but isn't a Baire space. Then by Lemma 2, there is some countable collection $\{A_n\}_{n=1}^\infty$ of closed nowhere-dense subsets of $X$ whose union doesn't have empty interior--meaning there is some non-empty open $U$ such that $$U\subseteq\bigcup_{n=1}^\infty A_n.$$ Now, in particular, we can take a non-empty open set $V$ such that $\overline V\subseteq U.$ (Why?) Then $$\overline V\subseteq\bigcup_{n=1}^\infty A_n,$$ so $$\overline V=\overline V\cap\left(\bigcup_{n=1}^\infty A_n\right)=\bigcup_{n=1}^\infty(\overline V\cap A_n).\tag{#}$$ But $\overline V$ is a non-empty complete metric space by Lemma 3, so by BCT, $\overline V$ is not a countable union of closed nowhere-dense subsets of $\overline V$. But each $\overline V\cap A_n$ is closed and nowhere-dense in $\overline V$ (why?), so $(\#)$ gives us the desired contradiction. $\Box$
It's certainly true that notions like meagre vs second category are only really "meaningful" in nice complete (or locally compact Hausdorff spaces, which are also Baire) spaces. There is another analogy one can consider too: the null sets (in $\Bbb R$) in the Lebesgue measure. This is a $\sigma$-ideal of "small" subsets of $\Bbb R$. One can think of the meagre sets as the smallest $\sigma$-ideal of the reals that contains all nowhere dense subsets, so all small sets wrt the topology (they indeed do not contain open sets, their complement is even dense! but a set like the irrationals in the reals (often denoted $\Bbb P$) also contains no open sets and has dense complement but is second category, so not meagre, so meagre-ness is more subtle, it's more $\Bbb Q$-like and less $\Bbb P$-like.
If you're interested in this stuff, there is an accessible short book by Oxtoby called "Measure and Category", in which he explores the similarities and analogies of these notions of meagre-ness vs measure $0$; both are important in analysis, but there are some subtle and nice differences too. A (classic) book to keep in mind, IMO.
Best Answer
Firstly, a minor adjustment: you can assume WLOG that the $A_n$ are not only closed, but increasing too: $A_n \subseteq A_{n+1}$ for all $n$. This is because finite unions of closed nowhere dense subsets are still closed nowhere dense and we can replace $A_n$ by $A'_n:=\bigcup_{i=1}^n A_i$ and we keep the same union and gain increasingness and so decreasingness of the $B_n$.
Secondly, make sure that the balls you're using are the closed balls of radii $< \frac{1}{n}$ at stage $n$ and ensure that each of closed balls lies inside the previous closed ball. Then the Cantor theorem on decreasing closed sets with diameters going to $0$ can be used and we have a point in the intersection (appealing to the completeness of $(X,d)$) and as each ball at stage $n$ avoids $A_n$, a point in the intersection cannot be in any $A_n$, which gives the contradiction we need. So the contradiction then comes from using completeness, not from refuting completeness by finding a wrong Cauchy sequence.
Cantor's theorem is the complete metric variant found here.