Covering the sphere with connected closed sets I am given the following question .
Can $S^{2}$ be covered by finite number of connected closed sets $A_{1},...,A_{n}$ such that

*

*Each has diameter less than a given $\epsilon$ .


*$\cup A_{n}= S^{2}$


*$\displaystyle A_{k}\cap\bigg(\bigcup_{i=1}^{k-1}A_{i}\bigg)$ is connected for all $k\leq n$ ?
My attempt:-
My first thought was to view it as a standard classic football and prove it for pentagons and hexagons. But then I realized that it is not very simple as ordering the sets such that property $3$ becomes a problem and moreover the pentagons and hexagons have a curvature in this case.
So I thought of covering with small enough "curved" squares . But I am struggling to make the argument fully rigorous.
I view the height of the sphere which is $1$ as the interval $[0,1]$ and divide into $\frac{1}{n^{2}}=\frac{1}{m}$  many may pieces where $\epsilon>\frac{1}{n}$.
Now the surface area of the sphere between $0\leq h\leq \frac{1}{m}$ is just $\frac{2\pi}{m}$ . And so I divide this surface area into $m$ many "curved squares" of area $\frac{2\pi}{m^{2}}$ and stack them so as the property $3$ holds.
Now it is hard to find the exact diameter $=\sup\{d(x,y):\,x,y\in A_{i}\}$    . But as the area of the square is taken to be small enough , we can say that it is less than $\epsilon$. If not then we can again scale it to make it so.
Now we proceed inductively and cover the sphere up with such squares for each interval of heights $[\frac{k}{m},\frac{k+1}{m}]$ for $k=1,...,m-1$.
Now this seems intuitively okay but I am worried about the connectedness part as we move up a level in height and I can't seem to make it fully rigorous.
Can someone  explain to me if I am going wrong somewhere or can suggest me a better way of solving this ?
Any help is appreciated.
PS. This was asked in a Algebraic Topology course as a similar argument was used to lift a path $\alpha :[0,1]\to X$ given a space $Y$ and a covering map $p$ .

Look at a globe:

Meridians and latitude circles divide $S^2$ into closed triangular surface pieces (around north and south pole) and closed quadrangular surface pieces. Taking sufficiently many meridians and latitude circles ensures that all these pieces have diameter less than the given $ϵ$. Number the pieces as follows:

*

*Number the latitude circles from north to south by $L_1,\ldots, L_m$ and the meridians counterclockwise by $M_1,\ldots,M_n$. Set formally $M_{jn+i} = M_i$ for all $j \in \mathbb N$ and $i = 1,\ldots, n$.


*For $i=1,\ldots, n$ let $A_i$ be the triangular surface piece bounded by $M_i, M_{i+1}$ and $L_1$.


*For $j = 1,\ldots, m-1$ and $i = jn+1,\ldots,jn +n$ let $A_i$  be the quadrangular surface piece bounded by $M_{i}, M_{i+1}$ and $L_j, L_{j+1}$.


*For $i = mn+1,\ldots,mn +n$ let $A_i$ be the triangular surface piece bounded by $M_{i}, M_{i+1}$ and $L_m$.
This gives us $N = (m+1)n$ surface pieces. It is clear that $\displaystyle A_{k}\cap\bigg(\bigcup_{i=1}^{k-1}A_{i}\bigg)$ is connected for all $k\leq N$.
