$X, Y$ metric spaces, $ f: X \rightarrow Y $ a function such that $ d_{2} (f(x), f(y)) \leq K d_{1}(x,y)$ $ \Rightarrow $ $ f $ is continuous. If $ (X, d_{1}) $ y $ (Y,d_{2}) $  are metric spaces and $ f: (X, \tau_{d_{1}}) \rightarrow (Y,\tau_{d_{2}})  $ is a function such that $ d_{2} (f(x), f(y)) \leq K d_{1}(x,y)$ for any $ x, y \in X $, then $ f $ is continuous.
We will see that for each $ \varepsilon > 0 $ and $ x \in X $, there exists some $ \delta > 0 $ such that if $ y \in X $ and $ d_{1}(x,y) < \delta $, then $ d_{2}(f(x),f(y)) < \varepsilon $.
For fixed $ \varepsilon > 0 $ and $ x \in X $, take $ \delta = \varepsilon / K $. Therefore, if $ y \in X $ and $ d_{1}(x,y) < \delta $, then $ d_{2}(f(x),f(y)) \leq K d_{1}(y,x) < K\delta=\varepsilon,$ as desired.
Is this proof correct? Where are the induced topologies  $ \tau_{d_{1}} $ and $ \tau_{d_{2}} $ used?
 A: This is 99% correct.  I would only change a few words, only based on many years of reading and writing elementary proofs.

We will see that for each $ \varepsilon > 0 $ and $ x \in X $, there exists some $ \delta > 0 $ such that ...

I would write show instead of see.  Think of this as a conversation.  You, the prover, is going to show something.  I, the reader, will see it once you've shown it.  We [will] show is more often used before a statement is justified, and we see afterwards.

For fixed $ \varepsilon > 0 $ and $ x \in X $, take $ \delta = \varepsilon / K $.

I think a more idiomatic version of this would be “Given $\varepsilon > 0$ and $x \in X$, let [or maybe set] $\delta = \varepsilon/K$.”  Use given to echo for all in the definition.  I might even put $x\in X$ ahead of $\varepsilon > 0$, but that's just picky because that's the order they would come in the definition of continuous.  There's no logical difference in the order.

Therefore, if $ y \in X $ and $ d_{1}(x,y) < \delta $, ...

I find therefore weird here because that's usually used to indicate you've completed a proof, or a major part of it.  And we're just getting to the meat of the proof.  I would write “Then, if $y \in X$,“ or (again) “Given $y\in X$ with $d_1(x,y) < \delta$, ...”
Finally, at the end, I would add, “Therefore [here is a good place to use that!  You could also say "We see that...”], $f$ is continuous at $x$.  Since $x$ was an arbitrary point of $X$, $f$ is continuous.  $\Box$

You asked where the induced topologies were used.  The metric space definition of continuous is equivalent to the topological one ($U$ open in $Y$ $\implies$ $f^{-1}(U)$ open in $X$) when you use the metric topology.
