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.
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
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.