Show that  the following subset Ω of Euclidean space is open

Let $\mathbf{x} \in \Omega$ be an arbitrary point. We will show that there exists $\delta > 0$ such that $B(x; \delta)$, the ball of radius $\delta$ around $x$, is contained in $\Omega$. Since $\mathbf{x} \in \Omega$, by definition we have $\lVert \mathbf{x} - \mathbf{a} \rVert < r$. Now let $\delta = r - \lVert \mathbf{x} - \mathbf{a} \rVert$. Clearly, $\delta > 0$, since $\lVert \mathbf{x} - \mathbf{a} \rVert < r$. Now we claim that $B(x; \delta) \subseteq \Omega$.

To see, let $\mathbf{y} \in B(x; \delta)$. Then $\lVert \mathbf{y} - \mathbf{x} \rVert < \delta$. Thus, by the triangle inequality,
$$\lVert \mathbf{y} - \mathbf{a} \rVert \leq \lVert \mathbf{y} - \mathbf{x} \rVert  + \lVert \mathbf{x} - \mathbf{a} \rVert < \delta + \lVert \mathbf{x} - \mathbf{a} \rVert = r.$$
Thus, by definition of $\Omega$, we have $\mathbf{y} \in \Omega$. Thus $B(x; \delta) \subseteq \Omega$. Therefore, $\Omega$ is open.