Arbitrary unions and finite intersections of open sets are open, open balls are open and closed balls are closed

Claim 1: let $y \in B(x,r)$, so $d(x,y) < r$, and put $s := r - d(x,y) > 0$; for $z \in B(y,s)$ the triangle inequality gives $d(x,z) \le d(x,y) + d(y,z) < d(x,y) + s = r$, so $B(y,s) \subseteq B(x,r)$, and since $y$ was arbitrary $B(x,r)$ is open.

Claim 2: let $y \in \bigcup \mathcal{U}$, so $y \in U$ for some $U \in \mathcal{U}$; as $U$ is open there is $t > 0$ with $B(y,t) \subseteq U \subseteq \bigcup \mathcal{U}$, and since $y$ was arbitrary the union is open.

Claim 3: let $y \in U_0 \cap \dots \cap U_{n-1}$ and for each $k < n$ pick $t_k > 0$ with $B(y,t_k) \subseteq U_k$, which is possible because each $U_k$ is open and $y$ lies in it.

Claim 4: let $y \in X \setminus \bar B(x,r)$, so $d(x,y) > r$, and put $s := d(x,y) - r > 0$; for $z \in B(y,s)$ the reverse triangle inequality applied to the points $y, z, x$ gives $d(y,x) - d(z,x) \le d(y,z) < s$, hence $d(z,x) > d(y,x) - s = r$, so $d(x,z) > r$ by symmetry and $z \notin \bar B(x,r)$.

Since $n \ge 1$, the set $\{t_0, \dots, t_{n-1}\}$ is a nonempty finite set of reals, so $t := \min\{t_0, \dots, t_{n-1}\}$ exists, equals some $t_j$ and is therefore $> 0$, and satisfies $t \le t_k$ for every $k < n$.

Step 1.4 shows $B(y,s) \subseteq X \setminus \bar B(x,r)$ for the $y$ and $s$ chosen there, and $y$ was an arbitrary point of $X \setminus \bar B(x,r)$; hence $X \setminus \bar B(x,r)$ is open and $\bar B(x,r)$ is closed, which is claim 4.

By step 2.1, $B(y,t) \subseteq B(y,t_k) \subseteq U_k$ for every $k < n$, so $B(y,t) \subseteq U_0 \cap \dots \cap U_{n-1}$; as $y$ was arbitrary that intersection is open, which is claim 3.

Claims 1, 2, 3 and 4 are established by steps 1.1, 1.2, 3.1 and 2.2 respectively.