The key here is to work with the function $a \mapsto d(a, \Omega^c) := \inf\{\|a-x\|:x \in \Omega^c\}$, where it is easy to prove both that
$$\tag{1}\inf\{\|a-x\|: a \in \bar{A}, x \in \Omega^c\}= \inf\{d(a, \Omega^c): a \in \bar{A}\},$$
$$\tag{2} d(\cdot,\Omega^c) \in C(\bar{A})$$
Since $\Omega$ is an open set, for any $a \in \bar{A} \subset \Omega$ there exists $\delta_a > 0$ such that the open ball $B(a;\delta_a)$ is contained in $\Omega$. If $x \in \Omega^c$, then $x \notin B(a; \delta_a)$ and $\|a-x\| \geqslant \delta_a > 0$. This implies that $d(a, \Omega^c) \geqslant \delta_a > 0$.
Since $\bar{A}$ is closed and bounded and, hence, compact and the distance function is continuous, there exists $a^* \in \bar{A}$ such that
$$\inf\{\|a-x\|: a \in \bar{A}, x \in \Omega^c\}=\inf\{d(a, \Omega^c): a \in \bar{A}\}= d(a^*,\Omega^c) \geqslant \delta_{a^*} > 0$$
Proof of (2).
To prove that $a \mapsto d(a,\Omega^c)$ is continuous note that for $a_1,a_2 \in \bar{A}$ and any $x \in \Omega^c$, we have, by the reverse triangle inequality
$$d(a_1,\Omega^c)- \|a_1-a_2\| \leqslant \|a_1-x\| - \|a_1-a_2\| \leqslant\|a_2-x\|$$
Thus, $d(a_1,\Omega^c)- \|a_1-a_2\| \leqslant \inf \{\|a_2 - x\|: x \in \Omega^c\} = d(a_2,\Omega^c)$ and after rearranging,
$$d(a_1,\Omega^c) - d(a_2, \Omega^c) \leqslant \|a_1 - a_2\|$$
Whence, switching $a_1$ and $a_2$ leads to
$$|d(a_1,\Omega^c) - d(a_2, \Omega^c)| \leqslant \|a_1 - a_2\|,$$
and it follows that $a \mapsto d(a, \Omega^c)$ is continuous.
Best Answer
$\frac 1 {\lambda +\mu} (x+y) =\frac {\lambda} {\lambda +\mu} \frac 1 {\lambda }x+ \frac {\mu} {\lambda +\mu} \frac 1 {\mu }y \in U$ whenever $ \frac 1 {\lambda }x \in U$ and $ \frac 1 {\mu }y \in U$. Can you complete the proof of triangle inequality, given this information?