Let $X$ be a separable complete metric Space. Then for each $n \in \mathbb{N}$, $X$ has a finite or countably infinite partition $\{A_{n,k}\}_k$ into nonempty Borel subsets $A_{n,k}$ of diameter at most $\frac{1}{n}$.
my Proof: Let $\{x_1,x_2,\dots\}$ be a countable, dense subset of X. We define $A_{n,1} := \{x \in X \mid d(x,x_1)\leq \frac{1}{n}\}$ and $A_{n,k+1}:= \{x \in X \mid d(x,x_k) \leq \frac{1}{n}\} \setminus \: \bigcup\limits_{i=1}^{k} A_{n,i}$.
Then the $A_{n,k}$ are disjoint Borelsets in $X$, $diam(A_{n,k})\leq \frac{1}{n}$ and from the separability of the $\{x_1,x_2,\dots\}$ it follows that $X= \bigcup\limits_{k \geq 1}\{x \in X\mid d(x,x_k) \leq \frac{1}{n}\}=\bigcup\limits_{k \geq 1}A_{n,k}$. So we can choose all $A_{n,k}$ that are not empty and we got our partition.
The idea is okay.
But you should go for e.g. $A_{n,1}=\{x\in X\mid d(x,x_1)<\frac1{2n}\}$ ensuring that for $x,y\in A_{n,1}$ we have: $$d(x,y)\leq d(x,x_k)+d(x_k,y)\leq\frac1{2n}+\frac1{2n}=\frac1{n}$$
Then consequently: $$\mathsf{diam}(A_{n,1})=\sup\{d(x,y)\mid x,y\in A_{n,1}\}\leq\frac1n$$
Also this must be adapted in the definition of the $A_{n,k}$ with $k>1$.
Small thing of course, and easily overlooked.
Further if I would be the one who gave you this question as homework then I would forgive you the first mistake but would like to see a real proof of the fact that the sets cover $X$:
Let $x\in X$ and let $D=\{x_1,x_2,\dots\}$ denote the countable and dense set. Then the density assures that $\{y\in D\mid d(x,y)\leq\frac1{2n}\}\neq\varnothing$. Now let $k$ be the unique positive integer with $d(x,x_k)\leq\frac1{2n}$ and $d(x,x_i)>\frac1{2n}$ for every $i\in\{1,\dots,k-1\}$. Then $x\in A_{n,k}$ and we are ready.