Baire category theorem and countable union of closed meagre sets in complete metric space

Question

In my functional analysis course Baire category theorem is the following: Complete metric spaces are of second category. The following was stated as Baire Category theorem's consequence: Suppose $X$ is a complete metric space. Suppose $c_1,...,c_n,...$ are closed meagre sets. Then $c = \cup_{n=1}^{\infty}c_n$ is also a meagre set. I can prove this in the following way. Suppose $c$ is not meagre and thus contains a closed ball. Suppose that ball is $B_0 = B[x_0, r_0]$. Then there exists a ball $B_1 = B[x_1,r_1] \subset B_0, r_1 < r_0/2$ such that $B_1 \cap c_1 = \emptyset$. Then $\exists B_2 = B[x_2,r_2] \subset B_1 \subset B_0, r_2<r_1/2<r_0/2^2$ such that $B_2 \cap c_1 = \emptyset, B_2 \cap c_2 = \emptyset$. In this way we get a sequence $B_n = B[x_n,r_n]$ such that $B_{n+1} \subset B_n$ and $r_n < r_0/2^n$. $X$ is complete so $\cap_{n=0}^{\infty} B_n = {a}$ where $ a \in X, a\notin c$. But this is a contradiction which is the end of the proof. Now, this is very similar to how Baire category theorem is proven. But it is listed as a consequence so I don't know how this result could be derived from the initial theorem statement. Also, I've noticed that this theorem is stated in different ways which are (I assume) equivalent and the above statements are two of those. While I can see clear correlation between the other possible ways to state the theorem, I don't see direct correlation between the above two.