Could you give me an example of a complete metric space with covering dimension $> n$ all of which closed separable subsets have covering dimension $\le n$?
The question closely related to this one.
The question closely related to this one. 


There are completely metrizable spaces $X$ with $\dim(X)=1$ and $\mathrm{ind}(X)=0$, where $\dim$ denotes the covering dimension and $\mathrm{ind}$ the small inductive dimension. The first such example is due to P. Roy (see "nonequality of dimensions for metric spaces", TAMS 1968), but there are many others. My favourite is due to E.K. van Douwen (see "The small inductive dimension can be raised by the adjunction of a single point", Indag.Math. 1973). Note that if $Y$ is a separable (not necessarily closed) subspace of such an $X$ then $\dim(Y)=0$ since $\dim$ and $\mathrm{ind}$ coincide for separable metrizable spaces. So any such $X$ satisfies what you want for $n=0$. 


@: $\varepsilon\delta$: $X$ isn't the the topological sum of real lines. You consider this "more shorty paths" logic: For move in a vertical line you go on the line. But for go from $(x, y)$ to $x', y')$ ($x\neq x'$) you go (vertically) from $(x, y)$ to $(0, x)$ then go (horizontally) to $(x', 0)$ then go (vertically) to $(x', y'). These "more short paths" distance describe the metric of $X$. Now a open ball centered in $(x, 0)$ and radius $\epsilon$ is a triangle isosceles, with the top right corner (the half of a square), its hight is $\epsilon$, and its base (intersection by the $x$axis) is a open interval of lenght $2\epsilon$ (centered in $x$), and the intersection by {$(x, y)\in R^2  y>0 $} is a open euclidean triangle. 


Let $X$ the river metric on the positive plane $(R^+)^2$: $d((x,y),(u,v)) = yv$ if $x=u$ and $y + v + xu$ otherwise. (the xaxis is the river, there are othogonal paths in which we can move "as in R", while otherwise we have to go via the river first). Let $v_x$ the vertical line by orizonatal quote $x$, then the subspace topology of $v_x\setminus$ {$(x,0)$} is the euclidean topology, but any open that containing a point of $r_0$ the the bottom horizontal line is nonseparable (involving a no countable vertical segments) we call a such open a "bopen" . THen any covering $\mathcal{U}$ of $X$ as order 2, i.e. there exixt almost 3 elemets by noempty intersection: Considering a bopen $U\in \mathcal{U}$, exist a point $x\in r_o \cap (Cl(U)\setminus U)$ (i.e. in its boundary in $r_0$) then there exixt another bopen $V\in \mathcal{U}$ containing $x$, and $U\cap V$ is a bopen. The sets $U$, $V$ and $U\cap V$ are also open in the euclidean topology and we can assume also connected and containing its own projection on $r_0$, then because $R^2$ has dimension 2 follow that exist a point $(x', y)\notin U\cup V$ such that any open set containing $(x', y) $ intersect $U\cap V$ (otherwise we can make a cover of order 2, then any refinement has order 2). Then the element $F\in \mathcal{U}$ containing $(x', y)$ (like a open interval in the $v_{x'}$) meet also $U\cap V$ : IF NO,the $sup$extrem of the quote that $U\cap V$ can reachedin in the $v_{x'}$ vertical is minor of $y$. Any (open ball) $B_\epsilon(x)$ with $x\in r_0$ is like a halphsquare triangle then if this is included in $U\cap V$ it dont meet the follow open part $S\subset R^2$: Considering the two halph line by base in $x',y$ at right by $45°$ pendence and $45°$ at left (like the graph of $y=x$ traslated from origin to $(x', y)$) and consider the superior part $S$ of plane $R^2$ these halphline cut off, by these halph line included. but then $U\cap V$ being union of (open) balls dont meet $S$, then exixt a (euclidean) open neighbord of $(x', y)$ that dont meet $U\cap V$ . 

