Question: Note that $X$ is a metric space and $f:(X,\rho)\rightarrow (X,\sigma)$.

Disproving: Suppose $(x\_n)$ is a Cauchy sequence, $f$ is continuous and $(f(x\_n))$ is not a Cauchy sequence, then $\exists\epsilon'>0$ such that $\forall M\in\Bbb{N}$, there exists $i,j\geq M$ where $\sigma(f(x\_i),f(x\_j))>\epsilon'$. Also, knowing that $f$ is continuous, then we know that $\exists\delta'>0$ such that if $\rho(x\_i,y)<\delta'$, such that $y\in X$, then $\sigma(f(x\_i),f(y))<\epsilon'$. However, $y$ is not necessarily a term of the Cauchy sequence $(x\_n)$. Hence, it does not necessarily mean that $(f(x\_n))$ is also a Cauchy sequence.

Now, I just noticed after typing that this may get tagged as duplicate since the same question has been asked here: [If $X = \{x\_n:n \in \mathbb N\}$ is a cauchy sequence in a metric space $S$ and $f : S \rightarrow T$ is continuous , is $f(x\_n)$ a cauchy sequence?](https://math.stackexchange.com/questions/962953/if-x-x-nn-in-mathbb-n-is-a-cauchy-sequence-in-a-metric-space-s-and), but would my argument be valid in general? Thank you!

Answer: 
Take $$X=(0,+\infty)$$

$$x\_n=\frac 1 n$$

$$f:x\mapsto \frac{1}{x}$$

$$(x\_n) \text{ is Cauchy}$$

$$f \text{ is continuous}$$

$$f(x\_n)=n \text{ is not Cauchy}$$

Your statement will be true if $f$ is UNIFORMLY CONTINUOUS.
No, it is not correct. You start talking about $y$ without telling what $y$ is. But, above all, your approach only tells us why you think that the sequence $\bigl(f(x\_n)\bigr)\_{n\in\mathbb N}$ **might** not be a Cauchy sequence. The best approach, in this case, consists in providing a counterexample.