Proving $\lim_{x\to x_0}f(x)$ with epsilon delta definition I've asked the question below before with no answer, but I would like to stress that this time it is not a homework question (and also that I've spent hours trying to come up with a solution). 
This is the question:

Let f be a function defined around $x_o$. 
  For every $\epsilon>0$ there's some $\delta>0$ such that if $0<|x-x_0|<\delta$ and $0<|y-x_0|<\delta$ then $|f(x)-f(y)|<\epsilon$.

And what's needed to be proven is that $\lim_{x\to x_0}f(x)$ exists.
I've been told that there are two ways to do so: One is quite easy and requires Cauchy sequences (I haven't learned sequences yet, but I think I'll look it up sometime soon and try to solve it this way).
The second way is a direct way, which I've been told is cumbersome and unrecommended, but since this is the way I tried solving it so far, I am really curious as to how the proof goes and this is the way I'm asking about. I tried applying all kinds of inequalities but with no success.
Even a little hint/direction would be swell. Thank you in advance.
 A: The following proof is essentially taken from Zorich, Mathematical analysis I.
Let $\mathcal{B}$ denote the family of all deleted neighborhoods of $x_0$. For each $B \in \mathcal{B}$, define
$$
m_B = \inf_{x \in B} f(x), \quad M_B=\sup_{x \in B}f(x).
$$
Since
$$
m_{B_1} \leq m_{B_1 \cap B_2} \leq M_{B_1 \cap B_2} \leq M_{B_2}
$$
for any elements $B_1$ and $B_2$ of $\mathcal{B}$, the axiom of completeness of $\mathbb{R}$ implies the existence of some $A \in \mathbb{R}$ that separates the numerical sets $\{m_B\}_B$ and $\{M_B\}_B$ as $B$ ranges over $\mathcal{B}$. By assumption, given $\varepsilon>0$ there exists $B \in \mathcal{B}$ such that $M_B-m_b < \varepsilon$. Then $|f(x)-A| < \varepsilon$ whenever $x \in B$.
A: O.k.; here is a proof not using sequences:
For each $\epsilon>0$ choose a point $x_\epsilon$ with $0<|x_\epsilon-x_0|<\delta$, where $\delta=\delta(\epsilon)$ is described in the statement. Put
$$a_\epsilon:=f(x_\epsilon)-\epsilon\ ,\quad b_\epsilon:=f(x_\epsilon)+\epsilon\ .$$
Given $\epsilon$, $\epsilon'$ we have $a_\epsilon\leq f(x)\leq b_{\epsilon'}$ for any $x$ with $0<|x-x_0|<\min\{\delta,\delta'\}$. Keeping $\epsilon'$ fixed we see that $$a:=\sup_{\epsilon>0}a_\epsilon\leq b_{\epsilon'}\ ,$$ and since this is true for all $\epsilon'>0$ we conclude that $$b:=\inf_{\epsilon'>0} b_{\epsilon'}\geq a\ .$$
Now for each $\epsilon>0$ we have
$$f(x_\epsilon)-\epsilon=a_\epsilon\leq a\leq b\leq b_\epsilon=f(x_\epsilon)+\epsilon\ .\qquad(*)$$
Therefore $0\leq b-a\leq 2\epsilon$, and since this is true for all $\epsilon>0$ we conclude that $a=b$.
Consider now any $x$ with $0<|x-x_0|<\delta$. Then $f(x)$ lies also within the two bounds $f(x_\epsilon)\pm\epsilon$  given in $(*)$. Therefore $|f(x)-a|\leq 2\epsilon$. Since $\epsilon>0$ was arbitrary we have proven that $\lim_{x\to x_0} f(x)=a$.
