real analysis - Find the flaw in the given proof: about the limit of a sequence

I'm studying Real Analysis, and one problem gives me a trouble. The problem is as below:

Let $\{x_n\}$ be a sequence defined on $\mathbb{R}$ with $\displaystyle \lim_{n \to \infty} x_n = x$ for some $x \in \mathbb{R}$. Define a sequence $\{\sigma_n \}$ on $\mathbb{R}$ by

$$\sigma_n= \frac{1}{n} ( x_1 + x_2 + x_3 + \cdots + x_n)$$
Find the flaw of the proof below, which tries to show the claim.

Claim : The sequence $\{\sigma_n\}$ converges. In addition, $\displaystyle \lim_{n \to \infty} \sigma_n = x$.

Proof. Since $\displaystyle \lim_{n \to \infty} x_n = x$, for any $\epsilon >0$, there exists a natural number $N$ such that

$$n >N \quad \Rightarrow \quad \lvert x_n-x \rvert < \epsilon$$
Now fix $\epsilon >0$, and let $N_\epsilon$ be the natural number that satisfies the property above. Note that
$$\lvert \sigma_n -x \rvert = \lvert\frac{1}{n} (x_1 + x_2 + \cdots + x_n) - x\rvert \leq \frac{1}{n} (\lvert{x_1-x}\rvert + \cdots + \lvert{x_n-x}\rvert)$$
Now, for sufficiently large $n>N_\epsilon$, we can divide the term above as
$$\lvert \sigma_n -x \rvert = \frac{1}{n}(\lvert{x_1-x}\rvert + \lvert{x_2 - x}\rvert + \cdots + \lvert{x_{N_\epsilon}-x}\rvert)+\frac{1}{n}(\lvert{x_{N_\epsilon+1}-x}\rvert + \cdots + \lvert{x_n-x}\rvert)$$
Since the first term above has only finite constant terms,
$$\frac{1}{n}(\lvert{x_1-x}\rvert + \lvert{x_2 - x}\rvert + \cdots + \lvert{x_{N_\epsilon}-x}\rvert) \to 0 \quad \text{as} \quad n \to \infty$$
Now,
$$\lvert \sigma_n -x \rvert = \frac{1}{n}(\lvert{x_{N_\epsilon+1}-x}\rvert + \cdots + \lvert{x_n-x}\rvert) < \frac{1}{n} \times \epsilon (n-N_\epsilon) \to \epsilon$$

as $n \to \infty$. Therefore $\displaystyle \lim_{n \to \infty} \sigma_n = x$.

I understand that there is some problem in the proof, but I cannot clearly explain the answer! I think the problem comes from finding the limit not at once, but calculating the parts first. Could somebody explain this to me plainly?