In a compact metric space, if we keep adding closed balls centered on boundary, do we always cover the entire space? Let $X$ be a compact connected metric space, and let $W_1=B(x,r)$ denote the closed metric ball centered at $x\in X$ with radius $r$. We recursively define $W_{k+1}=W_k \cup B(y,r)$, where $y$ denote a point on the boundary of $W_k$ and $B(y,r)$ is not contained in $W_k$. 
Is it true that there always exist $n>0$ such that $W_n=X$ for some $n$? 
If not, is there any additional requirement that can make this true?
 A: It is true.
Let us explain a little more precisely how the $W_k$ are defined recursively.
By $U(x,r)$ we denote the open metric ball with radius $r$ centered at $x$. Note that if $\emptyset \ne W \subsetneqq X$ is closed, then the boundary $\text{bd} W = W \setminus \text{int} W \ne \emptyset$ (otherwise $W = \text{int} W$, i.e. $W$ would be closed and open, contradicting the fact that $X$ is connected).
Actually we construct an increasing sequence of closed $W_k$ and a sequence $y_k$ of points such that $B(y_k,r) \subset  W_k$ and $y_{k+1} \in \text{bd} W_k$ if $W_k \ne X$.
Start with any $y_1$ and set $W_1 = B(y_1,r)$.
Assume $W_1,\dots, W_k$ and $y_1,\dots,y_k$ have been constructed.
If $W_k = X$, then $W_{k+1} = W_k$ and $y_{k+1} = y_k$. If $W_k \ne X$, choose any $y_{k+1} \in \text{bd} W_k$ and set $W_{k+1} = W_k \cup B(y_{k+1},r)$. Note that $B(y_{k+1},r)$ intersects $X \setminus W_k$, i.e. we get $W_k \subsetneqq W_{k+1}$.
The sequence $y_k$ has an accumulation point $y$. Hence we find $m < n$ such that $d(y_m,y), d(y_n,y) < r/2$. Hence $d(y_m,y_n) < r$ which implies that $y_n \in U(y_m,r) \subset B(y_m,r) \subset W_m \subset W_{n-1}$. We conclude $y_n \in \text{int} W_{n-1}$. Assume that $W_{n-1} \ne X$. Then by construction $y_n \in \text{bd} W_{n-1} =  W_{n-1} \setminus \text{int} W_{n-1}$ which is a contradiction.
Edited:
Andreas Blass' comment suggests to use the following fact which is true for any compact metric space $(X,d)$:
There does not exists an infinite sequence of points $y_k \in X$ such that $y_{k+1} \notin \bigcup_{i=1}^k U(y_i,r)$ for all $k$, or in other words such that $d(y_l,y_k) \ge r$ for all pairs $l,k$ with $l > k$.
The above construction is just a special case of this. Note that $W_k = \bigcup_{i=1}^k B(y_i,r)$ and hence $\text{int} W_k \supset\bigcup_{i=1}^k U(y_i,r)$. As long as $\text{bd} W_k \ne \emptyset$, we choose $y_{k+1} \in \text{bd} W_k$ which means in particular that $y_{k+1} \notin \text{int} W_k$ and hence $y_{k+1} \notin \bigcup_{i=1}^k U(y_i,r)$. We conclude that some $\text{bd} W_k = \emptyset$ which means that $W_k$ is closed an open, thus $W_k = X$ since $X$ is connected.
