Standard bounded metric induces same topology

Well, you can simply use the fact that $d$ and $d'$ agree on small balls. By the definition, $A\subset X$ is open iff for any $x\in A$ there exists $r>0$ such that $B(x,r)\subset A$.

Suppose that for some $x_0\in A$ and $r>0$ it holds that $B_{d'}(x_0,r)\subset A$. Then for any $q\in (0,r)$ it holds that $B_{d'}(x_0,q) \subset A$. In particular, it holds for $q = \min(\frac12,r)<1$ but then $$ B_{d'}(x_0,q) = B_d(x_0,q). $$ In similar lines you can show the converse. Equivalence of metric is sufficient for the equivalence of the induced topologies, but is not necessary.

In Topology by James Munkres, a different approach is taken:

Now we note that in any metric space, the collection of $\epsilon$-balls with $\epsilon<1$ forms a basis for the metric topology, for every basis element containing $x$ contains such an $\epsilon$-ball centered at $x$. It follows that $d$ and $\bar{d}$ induce the same topology on $X$, because the collection of $\epsilon$-balls with $\epsilon<1$ under these two metrics are the same collection. (page 122)

That last sentence was a bit confusing to me at first, so let me expand with what I think that means. Clearly, the topology induced by $\bar{d}$ is coarser than the topology induced by $d$. Now consider an open set $U$, an arbitrary element $y\in U$, and the ball $B_d(y,\epsilon)\subset U$ in the topology induced by $d$. We're given $\exists \epsilon$ such that $B_d(y,\epsilon)=\{b|d(b,y)<\epsilon\}\subset U$. If $\epsilon<1$ then $B_d(y,\epsilon)=B_\bar{d}(y,\epsilon)$, so clearly $B_\bar{d}(y,\epsilon)\subset U$. And if $\epsilon>1$, $B_\bar{d}(y,\epsilon)=\{b|d(b,y)<1\}\subset \{b|d(b,y)<\epsilon\}=B_d(y,\epsilon)\subset U$. Hence, the topology induced by $\bar{d}$ is also finer than the topology induced by $d$. So they induce the same topology.

Of course I'm studying this now, so take the last paragraph with a pound of salt.