Please verify that my proof is correct.
Problem: Let $f\colon (0,1)\to\Bbb R$ be uniformly continous. Let $(x_n)_{n=1}^\infty$ be a sequence in (0,1). Suppose that $\lim_{n \to\infty} x_n=0$. Prove that $\lim_{n \to\infty}f(x_n)$ exists.
My solution:
Since $\lim_{n \to\infty} x_n$ exists and is equal to $0$, this means the sequence ${x_n}$ is convergent. We know that every convergent sequence of real numbers is cauchy. Therefore ${x_n}$ is a cauchy sequence. Since $f$ is uniformly continous on $(0,1)$, for any given $\epsilon>0$ , $\exists$$\delta>0$ such that for every pair of points $x,y\in(0,1)$ satisfying
$$|x-y|<\delta\to|f(x)-f(y)|<\epsilon $$
Since ${x_n}$ is a cauchy sequence, $\exists$$k$$\in\mathbb{N}$ such that
$$|x_n-x_m|<\delta\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\forall n,m$$
which follows that $\forall n,m>k$
$$|f(x_n)-f(x_m)|<\epsilon\:\:\:\:\text{in}\:\:\mathbb{R} \:\:\implies f(x_n) ~~\text{is a Cauchy sequence} $$
Since every Cauchy sequence of $\mathbb{R}$ is convergent.
Therefore, $f(x_n)$ is convergent. Hence $\lim_{n \to\infty}f(x_n)$ exists
$$$$ Does my proof look correct?
