The étalé space associated to the sheaf of continuous functions over $\mathbb{R} $? We consider the sheaf $C$ of continuous functions over the reals, that's it, for any open subset $U$ of the reals we put $C(U) $ as the set of all continuous functions from $U$ to $\mathbb {R} $. There is an equivalence between the category of sheaves over a topological space $X$ and the category of étalé space over $X$. So for this sheaf, there must be an étalé space $p:E\longrightarrow \mathbb {R} $ such that the sheaf of sections is isomorphic to this sheaf. My question is there a description of the space $E$? What properties does $E$ have? $E$ is Haussdorf, compact, separable, connected,...?
 A: Etale spaces of sheaves are rarely any sort of nice familiar space you're used to, and are instead usually giant monstrosities.  The etale space of a sheaf over $X$ is of course locally homeomorphic to $X$, but beyond that, it will typically be quite nasty.
So, to illustrate, let me answer the specific questions you asked for the etale space $E$ of the sheaf $C$ of continuous (real-valued) functions on $\mathbb{R}$.  First, $E$ is not Hausdorff.  For instance, let $f:\mathbb{R}\to\mathbb{R}$ be the zero function and let $g:\mathbb{R}\to\mathbb{R}$ be a continuous function such that $g(x)=0$ for all $x\leq 0$, all $x\geq 1$ and $x\in[\frac{1}{2n+1},\frac{1}{2n}]$ for each positive integer $n$ but $g$ is nonzero on $(\frac{1}{2n},\frac{1}{2n-1})$ for each positive integer $n$.  Let $f_0,g_0\in E$ be the germs of these functions at $0$.  Then $f_0\neq g_0$, since there is no neighborhood of $0$ on which $f$ and $g$ are identically equal.  However, $f_0$ and $g_0$ do not have disjoint neighborhoods.  Indeed, a basic open neighborhood of $f_0$ has the form $\{f_t:t\in(-\epsilon,\epsilon)\}$ for some $\epsilon>0$ (where $f_t$ is the germ of $f$ at $t$), and similarly for $g_0$.  Any two such neighborhoods of $f_0$ and $g_0$ intersect, since for any $t\in (\frac{1}{2n+1},\frac{1}{2n})$, $f_t=g_t$.
Since the projection map $p:E\to\mathbb{R}$ is surjective and $\mathbb{R}$ is not compact, $E$ is not compact either.  Of course, since $E$ is locally homeomorphic to $\mathbb{R}$, it is at least locally compact.
$E$ is not separable.  For instance, for each constant function $c$, the set $U_c$ of all germs of $c$ is open in $E$.  These sets $U_c$ are all disjoint, and there are uncountably many of them.  So any dense subset of $E$ must be uncountable, in order to contain a point from each $U_c$.
Finally, $E$ is path-connected.  Indeed, consider two germs $a,b\in E$.  If $p(a)\neq p(b)$, then we can find a continuous function $f:\mathbb{R}\to\mathbb{R}$ whose germ at $p(a)$ is $a$ and whose germ at $p(b)$ is $b$.  Indeed, just choose a function with germ $a$ and a function with germ $b$, shrink their domains to be disjoint closed neighborhoods of $p(a)$ and $p(b)$, and then extend continuously to a function on $\mathbb{R}$.  This function $f$ then is a global section of $C$, so the map $s:\mathbb{R}\to E$ sending $t\in \mathbb{R}$ to the germ of $f$ at $t$ is continuous.  This gives a path from $a$ to $b$.  This was all assuming that $p(a)\neq p(b)$, but if $p(a)=p(b)$, then we can just choose some $c$ with $p(c)\neq p(a)$ and then find paths from $c$ to $a$ and $b$ as above.
