Can every domain be exhausted by compact *connected* subsets? 
Let $D \subset \Bbb R^n$ be an open connected set. I would like to exhibit an increasing sequence of compact connected subsets of $D$ converging to $D$.

For example, for a ball we might take a sequence of closed ball inside it of radius $r-\frac 1n$.
User SteamyRoot showed that the usual closed exhaustion that generalizes the above example, given by $D_n = \{ x\in D: \mathrm{dist}(x, \partial D) \geq \frac 1n \}$ is not always connected (by considering the shape of eye-glasses), even before modifying it to be bounded. Considering an infinite sequence of glasses connected to one another in a row that are smaller and smaller shows that this will not be connected no matter how small we choose $n$.
We also cannot do this by taking the union of all closed squares of side length $2^{-n}$ contained entirely within $D$. This is again not connected because of the same counterexample (infinite glasses), even though this example does have such an exhaustion.

Fix a point $x_0 \in D$. For every $A$ subset of $D$ containing $x_0$ denote by $A'$ the component of $A$ containing $x_0$. Assume now we have $D_m$ an increasing sequence of open subsets of $D$ containing $x_0$ and with union $D$. Then $D_m'$ (the $x_0$ components)  also cover $D$. Indeed, consider $x \in D$. There exists a path $\gamma$ from $x_0$ to $x$ contained in $D$.  Since the $D_n$'s cover $D$ and $\gamma$ is compact there exists $n$ so that $\gamma \subset D_m$. Then $x\in D_m'$. 
Take now any increasing sequence of compacts $K_m$ whose interiors cover $D$ ( for instance 
$K_m= \{ x \in \mathbb{R}^n \ | \ ||x|| \le m \textrm { and } d(x, \mathbb{R}^n \backslash D ) \ge \frac{1}{m} \}$). Then
$L_m = K_m'$ will be an exhaustion of $D$ with connected compacts.  
$\bf{Added:}$ In the construction above, we can also take $L_m$ to the the closure of the $x_0$ connected component of $\mathring{K_m}$ ( the interior of $K_m$). The advantage is that the interior of $L_m$ is connected and its closure is $L_m$, so $L_m$ is better behaved. 
In general, if $L$ is a compact subset of the open set $D$, consider a covering of the full space with a lattice of $\epsilon$ cubes so that the diagonal of the cubes $\sqrt{n} \cdot \epsilon < d(L, \mathbb{R}^n \backslash D)$. Consider $\tilde L$ to be the union of all the cubes that intersect $L$. Then $\tilde L$ is a compact and $L \subset \tilde L \subset D$. If, moreover, $L$ is connected then $\tilde L$ is connected. 
If $L$ has connected interior $\mathring{L}$ and is the closure of its interior, then the same holds for $\tilde{L}$. So we can get  pretty well behaved compact exhaustions. 
