If continuity preserves convergence, and Cauchy sequences are convergent sequences, why do we need uniform continuity to preserve Cauchy sequences? In $\mathbb R$, all Cauchy sequences are convergent and all convergent sequences are Cauchy. So, why isn't continuity enough to preserve Cauchy sequences?
A function is continuous iff it preserves convergent sequences.
A sequence is convergent iff it is Cauchy.
So, why doesn't it follow that continuous functions preserve Cauchy sequences?
 A: why isn't continuity enough to preserve Cauchy sequences?
Because, in general, it is not given that the limit of the sequence belongs to the domain of the function.
The implication
$$\lim_{n\to\infty} x_n= x\quad \Longrightarrow\quad \lim_{n\to\infty} f(x_n)= f(x)$$
requires $x_n,x\in D(f)$. Thus, if $x\notin D(f)$, this argument does not work (a counterexample was given in the comments of your post). In fact, in this case we need an extra condition (see the first comment below).

Edit
Let us analyze your argument:

Claim: Continuous functions preserve Cauchy sequences.
Poof: Let $X$ be a subset of $\mathbb R$. Let $f:X\to\mathbb R$ be a continuous function. Let $(x_n)$ be a Cauchy sequence.
We want to show that $(f(x_n))$ is Cauchy.
  
  
*
  
*As a (real) sequence is convergent iff it is Cauchy, there exists $x_0\in\mathbb R$ such that
  $$\lim_{n\to\infty}x_n=x_0.\tag{$*$}$$
  
*As a function is continuous iff it preserves convergent sequences, there exists $L\in\mathbb R$ such that
  $$\lim_{n\to\infty}f(x_n)=L.$$
  
*As a (real) sequence is convergent iff it is Cauchy, we conclude that $(f(x_n))$ is Cauchy.

The problem is step 2 which, in general, does not follow from $(*)$. If $x_0$ belongs to $X$, then it is true (with $L=f(x_0)$) but in general we cannot guarantee the existence of $L$.
