I'm looking for feedback to the following proof. I'm most interested in knowing whether there is a more straightforward proof.
Let $d_1, d_2$ be two metrics on $X$ s.t. $\forall x, y\in X: d_1(x, y)\leq d_2(x, y)$. We will show that $\mathcal{T}_{d_1}\subset \mathcal{T}_{d_2}$, where the two topologies are topologies generated by the metrics $d_1, d_2$ on $X$. Thus let $U \in \mathcal{T}_{d_1}$ and $x \in U$. Then for some $r_1 > 0$ the ball $B_{d_1}(x, r_1)\subset U$. Suppose that $y \in B_{d_1}(x, r_1)$ and let $\epsilon > 0$ be arbitrary. Then for $r_2 := \max\{r_1, d_2(x, y) + \epsilon\}$ we have that $y \in B_{d_2}(x, r_2)$. Then if $B_{d_2}(x, r_2) \not\subset U$, it follows that for some $z \in X\setminus U$ we have that $d_2(x, z) < r_2$. But $r_1 < d_1(x, z) \leq d_2(x, z)$ by $B_{d_1}(x, r_1)\subset U$. Therefore, $d_2(x, z) < r_2 = \max\{r_1, d_2(x, y)\}$ violates the triangle inequality $d_2(x, z) \leq d_2(x, y) + d_2(y, z)$ as $\epsilon > 0$ is arbitrary. Thus $B_{d_2}(x, r_2) \subset U$ so that by definition $U \in \mathcal{T}_{d_2}$. Hence $\mathcal{T}_{d_1}\subset \mathcal{T}_{d_2}$.
It's much simpler:
Let $O$ be $d_1$-open. Let $x \in O$ be arbitrary.
Then there is some $r>0$ so that $B_{d_1}(x,r) \subseteq O$. By the inequality:
$$B_{d_2}(x,r) \subseteq B_{d_1}(x,r)\tag{1}$$
(if $y \in B_{d_2}(x,r)$, we have $d_2(x,y) < r$ and then also $d_1(x,y) \le d_2(x,y) < r$ so $y \in B_{d_1}(x,r)$)
From $(1)$ it follows that $B_{d_2}(x,r) \subseteq O$, and as $x \in O$ was arbitrary, by definition $O$ is $d_2$-open.
No triangle inequality etc....