Let
- $(M,d)$ be a metric space
- $\overline B_\delta(x)$ denote the closed $\delta$-ball of $x$ in $M$
- $\Lambda\subseteq M$ be open
Now, let $\varepsilon>\delta>0$, $$\Lambda_\varepsilon:=\left\{x\in M:d(x,\Lambda^c)>\varepsilon\right\}$$ and $K\subseteq\Lambda_\varepsilon$ be compact.
Trivially, $$K\subseteq\bigcup_{x\in K}\overline B_\delta(x)\subseteq\Lambda_\varepsilon\tag1$$ and hence (since $K$ is compact) there are $n\in\mathbb N$ and $x_1,\ldots,x_n\in K$ with $$K\subseteq\bigcup_{i=1}^n\overline B_\delta(x_i)\tag2.$$
However, are we able to choose $x_1,\ldots,x_n$ such that $$\overline B_\delta(x)\subseteq\bigcup_{i=1}^n\overline B_\delta(x_i)\;\;\;\text{for all }x\in K\text?\tag3$$
In general not. Let $M = [0,1] \times [0,1] \subset \mathbb{R}^2$ endowed with the Euclidean metric and $\Lambda = [0,1] \times (0,1]$. Let $1 > \epsilon > \delta > 0$. Then $\Lambda_\epsilon = [0,1] \times (\epsilon,1]$. Let $K = [0,1] \times \lbrace 1 \rbrace$. We have $\bigcup_{x \in K} \overline{B}_\delta(x) = [0,1] \times [\delta,1]$. No finite subcover $\overline{B}_\delta(x_i)$ will cover this set.
This example can be generalized to $M = \mathbb{R}^d$ endowed with the Euclidean metric. Let $\Lambda = \mathbb{R}^{d-1} \times (0,\infty]$. Let $\epsilon > \delta > 0$. Then $\Lambda_\epsilon = \mathbb{R}^{d-1} \times (\epsilon,\infty]$. Let $K = [0,1] \times \lbrace (0,..,0,2\epsilon) \rbrace$. We have $\bigcup_{x \in K} \overline{B}_\delta(x) \supset [0,1] \times \lbrace (0,..,0) \rbrace \times [2\epsilon-\delta,2\epsilon]$. No finite subcover $\overline{B}_\delta(x_i)$ will cover this set. In fact, consider $\overline{B}_\delta(\xi)$ with $\xi = (t,0,...,0,2\epsilon)$. The point $(t,0,...,0,2\epsilon-\delta)$ is contained in no other $\overline{B}_\delta(\xi')$.
Concerning your modified question: If $\tilde{K}$ would exist, then each $x \in K$ must have a compact neighorhood. This means essentially that $M$ is locally compact (at least around $K$). Take $M = \mathbb{Q}^d$ as a subspace of $\mathbb{R}^d$ to get an example where you do not find the desired $\tilde{K}$ (for $K$ take a one-point subspace of $\Lambda_\epsilon$). In locally compact metric spaces (like $\mathbb{R}^d$) it is, however, true (at least for sufficiently small $\delta$):
Consider a compact $K \subset \Lambda_\epsilon$. Since $M$ is assumed to be locally compact, there exists an open neighborhood $U$ of $K$ such that $\overline{U}$ is compact and contained in $\Lambda_\epsilon$. Let $\delta = inf \lbrace d(x,y) \mid x \in K, y \in M-U \rbrace$. Since $K$ is compact and $M-U$ closed, we have $\delta > 0$. Therefore each $\overline{B}_\delta(x)$, $x \in K$, is contained in $\tilde{K} = \overline{U}$.