Let $(X,d)$ be a general metric space. Let $B=B(x,r)$ be the open ball with center $x$ and radius $r$. Let $B^{\varepsilon} := \{y \in X : d(y,B) = \inf_{z \in B} d(y,z) < \varepsilon\}$ be its $\varepsilon$-neighborhood. Is it true that $B(x,r+\varepsilon) = B^{\varepsilon}$?
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The answer is no.
Consider the subspace $X = \mathbb{R}\times\{0\}\cup \{(0,1)\}$ of the plane $\mathbb{R}^2$ (with the euclidean metric). Let $B = B((0,0),1) \in X$. Then $(0,1) \notin B$. Furthermore, the distance between $(0,1)$ and any interval in $\mathbb{R}\times\{0\}$ containing the center $(0,0)$ is $1$. Hence, $(0,1) \notin B^{\varepsilon}$ if $\varepsilon \leq 1$ by definition of the $\varepsilon$-neighborhood. However, $(0,1) \in B((0,0), 1 + \varepsilon)$ by the euclidean metric.
Note that even if we considered the surface of a rectangle in $\mathbb{R}^2$, that is, a complete, compact, connected, path connected metric space with the Heine Borel property, we would still be able to replicate the same problem.
To assure that we can get such equality we need a Length Structure in $X$ with $d$ intrinsic to the Length functional of the structure. In that way we can use the $\varepsilon$-mid point property of the space to give a simple proof that the equality holds.
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