Why does the support of measure on $\mathbb{R}^n$ exist? 
DEFINITION : The support of a measure on $\mathbb{R}^n$, written spt $\mu$, is the smallest closed set such that $\mu(\mathbb{R}^n \setminus X)=0$.

Why does this set exist?
 A: It exists because the topology of $\mathbb{R}^n$ is second countable$^1$.
By definition, the complement of the support of $\mu$ is the largest open set $U$ with $\mu(U) = 0$. Showing the existence of that set, and of the support of $\mu$ are equivalent.
So consider the family $\mathscr{V} = \{ V\subset \mathbb{R}^n : V\text{ open}, \mu(V) = 0\}$.
Then $U = \bigcup \mathscr{V}$ is an open set, and $U$ contains every open set with measure $0$. It remains to see that indeed $\mu(U) = 0$. That is where second-countability comes into play: Since the topology is second-countable, $U$ is in fact the union of a countable subfamily $\mathscr{V}_0$ of $\mathscr{V}$, and hence
$$\mu(U) = \mu\left(\bigcup_{V\in\mathscr{V}_0} V\right) \leqslant \sum_{V\in\mathscr{V}_0}\mu(V) = 0.$$
Note that we don't need to demand that $\mu$ be a Borel measure, there is always at least one open measurable set with measure $0$, namely $\varnothing$, and since only measurable open sets occur in $\mathscr{V}$, it follows that $U$ is measurable as a countable union of measurable (open) sets.
In general topological or metric spaces, there need not exist a smallest closed set whose complement has measure $0$:
Let $Y$ be an uncountable set with the discrete topology, and let
$$\mu(A) = \begin{cases} 0 &, A\text{ is at most countable}\\ \infty &, A\text{ is uncountable}.\end{cases}$$
Admittedly not the most interesting or useful measure, but it serves well as a counterexample to several things.
Then $\mu$ has no support (if the support is defined as the smallest closed set whose complement has measure $0$), since for every (closed) subset $S\subset Y$ with $\mu(Y\setminus S) = 0$ there is a proper subset $T\subsetneqq S$ with $\mu(Y\setminus T) = 0$, e.g. $T = S\setminus \{s\}$ will do for any $s\in S$.

$^1$ A separable metric space is second countable: Let $X$ be a separable metric space, and $D = \{x_n : n\in \mathbb{N}\}$ a countable dense subset. Then $$\mathscr{B} = \{B_{2^{-k}}(x_n) : n,k\in\mathbb{N}\}$$
is a countable basis of the topology of $X$.
Since $\mathbb{N}\times\mathbb{N}$ is countable, the family $\mathscr{B}$ is countable. To see it is a basis of the topology, we must see that for every open set $U$ and every $x\in U$ there is a $B\in\mathscr{B}$ with $x \in B \subset U$. But if $U$ is open and $x\in U$, there is an $\varepsilon > 0$ such that $B_\varepsilon(x) \subset U$. Choose $k\in\mathbb{N}$ such that $2^{1-k} < \varepsilon$. Since $D$ is dense, there is an $n\in\mathbb{N}$ with $x_n \in B_{2^{-k}}(x)$. Then by symmetry of the metric we have $x\in B_{2^{-k}}(x_n)$, and by the triangle inequality $B_{2^{-k}}(x_n) \subset B_{2^{1-k}}(x) \subset B_\varepsilon(x) \subset U$.
Thus $\mathscr{B}$ is a basis of the topology.
For $\mathbb{R}^n$, a canonical choice for the countable dense subset is $\mathbb{Q}^n$.
A: Is the intersection of all the closed sets such that... And can be empty.
A: $${\rm spt } \mu = \bigcap\left\{S \colon S \text{ is closed and } \mu(S^c)=0\right\}$$
exists, because $\mathbb{R}^n$ can be an example of a set $S$.
