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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

e W \subsetneqq X$ is closed, then the boundary $\text{bd} W = W \setminus \text{int} W

e \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

e 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

e 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}

e 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}

otin \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

e \emptyset$, we choose $y_{k+1} \in \text{bd} W_k$ which means in particular that $y_{k+1}

otin \text{int} W_k$ and hence $y_{k+1}

otin \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.