Upper semicontinuous function and equivalent statements Problem
Let $f:\mathbb R^n \to \overline{\mathbb R}$, then the following statements are equivalent:
(1) $f$ is upper semicontinuous;
(2) for every $t \in \overline{\mathbb R}$, $\{x \in \mathbb R^n: f(x)<t\}$ is open;
(3) for every $t \in \overline{\mathbb R}$, $\{x \in \mathbb R^n: f(x)\ge t\}$ is closed. 
I am trying to show the equivalence of the statements
It is immediate that (2) and (3) are equivalent since $A=\{x \in \mathbb R^n: f(x)<t\}$ is open if and only if $A^c$ is closed, but $A^c=\{x \in \mathbb R^n: f(x)\ge t\}$.
I am having problems showing (1) is equivalent to (2).
Suppose $f$ is upper semicontinuous. Let $x_0 \in \mathbb R^n$, by hypothesis we have $\inf M'_{\delta}(x_0) \leq f(x_0)$, where $M'_{\delta}(x_0)=\sup\{f(x): 0<|x-x_0|<\delta\}$. 
Now, let $t \in \overline{\mathbb R}$. I want to show that the set $A$ is open ($A$ is the set defined in (2)). If $A$ is empty, there is nothing to prove. Suppose $A$ is not open, then there is $x_0 \in A$ such that for all $n \in \mathbb N$ there is $x_n \in B(x_0,\frac{1}{n})$ with $f(x_n) \geq t$. We have $f(x_0)<t\leq f(x_n)$. But this means $M'_{\frac{1}{n}}(x_0)>f(x_0)$. 
From there I think I should get to a contradiction, I got stuck at that point. 
I have no idea how to show (2) $\implies$ (1), I would appreciate some help and suggestions.
 A: 
But this means $M'_{\frac{1}{n}}(x_0)>f(x_0)$.

It means a bit more, and that bit more is what you need. It means that for all $x_0 \in A_t$, where $A_t := \{ x\in\mathbb{R}^n : f(x) < t\}$, that aren't interior points of $A_t$, you have $M'_{\frac{1}{n}}(x_0) \geqslant t$. And, by the monotonicity of $M'_\delta(x_0)$ with respect to $\delta$, it follows that $M'_\delta(x_0) \geqslant t$ for all $\delta > 0$, hence
$$\inf_{\delta > 0} M'_\delta(x_0) \geqslant t > f(x_0),$$
i.e. $f$ is not upper semicontinuous if there is a $t$ such that $A_t$ is not open.
And to see that (2) implies (1) [Remark: usually, condition (2) - or, less often, (3) - is the definition of upper semicontinuity; the characterisation in terms of the $\limsup$ - or, for lower semicontinuity, the $\liminf$ - is then a(n occasionally used) proposition], pick any $x_0\in\mathbb{R}^n$, and note that for every $t > f(x_0)$, the set $A_t = \{ x \in \mathbb{R}^n : f(x) < t\}$ is a neighbourhood of $x_0$, so $\{ x : \lvert x-x_0\rvert < \delta\} \subset A_t$, and $M'_\delta(x_0) \leqslant t$ for all small enough $\delta > 0$. Therefore we have
$$\inf_{\delta > 0} M'_\delta(x_0) \leqslant t.$$
Since that holds for all $t > f(x_0)$, it follows that
$$\inf_{\delta > 0} M'_\delta(x_0) \leqslant f(x_0).$$
