Dimension of subspace is greater then dimension of space If $X$ is a topological space then it's (covering) dimension is defined as a minimal number $n$ such that for every finite open cover $\{U\}$ of $X$ there is a finite open cover $\{V\}$ of $X$ that refines $\{U\}$ and such that every point $x \in X$ is contained in no more than $n+1$ set of $\{V\}$.
If $X$ has a dimension $n$ and $F$ is a closed subspace of $X$ then $\dim F \leqslant n$. It is an easy exercise: if $\{U\}$ is an open cover of $F$ then $\{U\}$ and $F^c$ give and open cover of $X$ and we can use the definition of dimension for $X$ to obtain $\dim F \leqslant \dim X$.
But I didn't find in any book that it is true in general. Also I didn't find a counterexample.   Is it possible that a subspace $Y$ of $X$ has a dimension greater then dimension of $X$? If it is possible it is very interesting to look at such example.
 A: There are examples of T$_{3 \, 1/2}$-spaces $X$ and closed $F \subseteq X$ with $\mathrm{dim}(F) > \mathrm{dim}(X)$.
Problem 7.4.6 (p.419) of Engelking's General Topology (1989 ed.) asks for an example of such a space.  This example is therein credited to

Yu. M. Smirnov, On the dimension of proximity spaces (Russian), Mat. Sb. N.S. 38 (1956), 283–302 MR0082095 (English translation Amer. Math. Soc. Transl. (2) 21, 1962, 1–20.  MR150728).

For non-closed subsets, it is a bit easier (though I will again steal from Engelking).
Take a zero-dimensional not strongly zero-dimensional space $X$ (e.g., Dowker's Example, 6.2.20 in Engelking).  This space has a zero-dimensional compactification $\gamma X$ (Corollary 6.2.17) and every compact — in fact Lindelöf — zero-dimensional space is strongly zero-dimensional (Theorem 6.2.7).  Therefore $ \mathrm{dim} ( \gamma X ) = 0 < \mathrm{dim} (X)$.

Added: I see now that the definition of the covering dimension given applies only to normal spaces.  As stated in the OP, it is a theorem that closed subspaces of normal spaces cannot have larger covering dimension.
For the second example the space $X$ from Dowker's Example is normal, and clearly so, too, is its zero-dimensional compactification $\gamma X$, and so this would yield an example of a normal subspace of a normal space with strictly larger covering dimension.
