Dirac measure proof $ \mu B(\mathbb{R^n}) \rightarrow \{0,1\} $ is measure on Borel-sigma-algebra.
It should hold: $$ \mu(\mathbb{R^n})=1$$
I want to show: $ \exists x \in \mathbb{R^n} $, such that $ \mu=\mu_x$
$\mu_x$ is shall be the dirac measure.
My idea:
It holds: $$ \mathbb {R^n} = \cup_{n \in N} [-j,j)^n$$, that means:
$\exists k \in \mathbb{N} \  \mu([-k,k)^n) =1$
Now I can divide this cuboid into $2^n$  little cuboids.
There exists a little cuboid $Q_k$, such that $\mu(Q_k)=1$.
Furthermore the cuboids are compact. (How can I justify that?)
Then $ \bigcap_k Q_k =\{x\} $
Because $\mu$ is continous, it holds: $ \mu(\bigcap_k Q_k) =\lim_{k \rightarrow \infty}  \mu[Q_k)$ 
How can I go on from that?
 A: Your general plan to use a nest of cubes to trap the mass is a good one but its execution is flawed.
Here is how I would edit your idea.  First, construct an increasing nest of compact cubes $B_m=[-m,m]^n\in\mathbb R^n$.  (Compact because they are closed and bounded.) Since their union is all of $\mathbb R^n$, we know $$1=\mu(\mathbb R^n)\le\sum_{m\ge1}\mu(B_m).$$ For each $m$ we have $\mu(B_m)=0$ or $\mu(B_m)=1$, and so $\mu(B_m)=1$ for some $m$, for otherwise all the terms in this series would vanish.  For that $m$, inductively  subdivide  $B_m$, to obtain a decreasing nested family of compact cubes $Q_0, Q_1,\ldots$, where $Q_0=B_m$, and for which  $\mu(Q_k)=1$ for all $k$.    Starting with $k=0$ and $Q_0=B_m$, construct   $Q_{k+1}$  from $Q_k$ as follows.  Divide $Q_k$ into $2^n$ compact subcubes $S_{i,k}$, each of half the side width as $Q_k$, which I suppose works out to $2m2^{-k}$.  We know that $\mu(Q_k)\le \sum_i\mu(S_{i,k})$.  By the inductive hypothesis, $\mu(Q_k)=1$ so at least one of the summands $\mu(S_{i,k})$ is non-zero and hence equal to $1$.  Let $j$ be such that $\mu(S_{j,k})=1$, and then let $Q_{k+1}=S_{j,k};$ since 
$Q_{k+1}$ is just our new name for $Q_{k+1}$, we know $\mu(Q_{k+1})=1$.
(Here the subdivision is like $[0,1]=[0,1/2]\cup[1/2,1]$ and so on, where some points end up in both pieces.)
The rest of your argument is OK: the $Q_k$ are a decreasing sequence of  sets converging to the non-empty set $Q=\bigcap Q_k$, so $\mu(Q)=\lim_k \mu(Q_k) =1$.  Since the side lengths of $Q_k$ tend to $0$, the set $Q$ is actually a singleton.
A: Another, and I think simpler, way to show this is to look at the functions $$f_i(t)=\mu\left(\left\{x=(x_1,\ldots,x_n)\in \mathbb R^n : x_i\le t\right\}\right)$$
for $i=1,\ldots,n$. 
Each $f_i$ non-decreasing and for all $t\in\mathbb R$, $f_i(t)=0$ or $=1$.
By the continuity of $\mu$ from below, $\lim_{t\to\infty}f_i(t) = \mu(\mathbb R^n)=1$. 
By continuity of $\mu$ from above, $\lim_{t\to-\infty} f_i(t)=\mu(\phi)=0$.  Therefore  $f_i$ has a jump discontinuity at (say) $a_i$.  That is to say, $\mu$ concentrates its mass on the hyperplane $x_i=a_i$.
So $\mu$ concentrates all its mass on the intersection of these hyperplanes, that is to say, on the point $a=(a_1,\ldots,a_n)$.  
