Prove if $f$ is continuous at $x_0$ and $f(x_0)>\mu$, then $f(x)>\mu ,\forall x$ in some neighborhood of $x_0$ 
Prove if $f$ is continuous at $x_0$ and $f(x_0)>\mu$, then $f(x)>\mu ,\forall x$ in some neighborhood of $x_0$

My attempt:
$f$ is continuous implies:
$$\forall \epsilon>0, \exists \delta >0 \text{ s.t.} |f(x)-f(x_0)|<\epsilon \forall |x-x_0|<\delta$$
$$\epsilon \geq |f(x)-f(x_0)|=|f(x_0)-f(x)|\geq |f(x_0)|-|f(x)|\geq \mu-|f(x)|$$
i am not sure how to proceed?
 A: Hint:
Set $\epsilon = \frac{|f(x_0)-\mu|}{2}$
A: Without $ \epsilon - \delta$: assume that such a neighborhood does not exist. Then we get for each $n \in \mathbb N$ a point $x_n$ with
$|x_0-x_n|<1/n$ and $f(x_n) \le \mu$.
Since $x_n \to x_0$, we have $f(x_n) \to f(x_0)$. This gives the contradiction $f(x_0) \le \mu$.
A: Let's start by a remark, I think this wording is preferable as slightly clearer than rejecting the $\forall$ quantifier at the end.
$\forall \epsilon>0,\ \exists \delta >0\ \text{ s.t. } |x-x_0|<\delta\implies|f(x)-f(x_0)|<\epsilon$
Here the quantifier is even silenced and made implicit.

Another remark is that later you wrote :
$|f(x_0)-f(x)|\ge |f(x_0)|-|f(x)|\quad$   this is wrong
It should be $|f(x_0)-f(x)|\ge \bigg||f(x_0)|-|f(x)|\bigg|$ because you know nothing about the monotonicity of $f$.

But it is simpler to just expand the continuity inequality and select the lower bound part:
$-\epsilon<f(x)-f(x_0)<\epsilon\implies f(x)>f(x_0)-\epsilon\quad$  is really what's we are interested in.
We can rewrite it : $$f(x)>\mu+\underbrace{(f(x_0)-\mu)-\epsilon}_{\phantom{a}\\\text{we have to make this positive}}$$
Thus selecting $\epsilon=\frac 12(f(x_0)-\mu)>0$ appears now natural to prove $f(x)>\mu$.

You don't need $\frac 12$ you just need to make the quantity I specified positive. 
For instance if you take $\epsilon=\frac {3}{17}(f(x_0)-\mu)$ then $f(x)>\mu+\underbrace{\frac{14}{17}(f(x_0)-\mu)}_{>0}$ so $f(x)>\mu$. 
We just choose $\frac 12$ since it is by far the simplest. 
But generally $\epsilon=\lambda(f(x_0)-\mu)$ for $0<\lambda<1$ would work, not only fractions.
