Can Lipschitz maps increase the Lebesgue dimension ?

The answer to your first question is No. It is natural, as Lebesgue dimension is a topological invariant.

There is a very simple example, which took me however a little time to figure out. Consider the Cantor set obtained by recursively replacing a square by four squares of $1/4$-th side length, one at each corner, endowed with the metric inherited from the Euclidean one. There is an orthogonal projection that maps this Cantor set onto a segment, see

(source)

Such a map is $1$-Lipschitz, but a segment has Lebesgue dimension $1$ while a Cantor set has Lebesgue dimension $0$.

What you need to rule out Lipschitz space-filling curves is for example Hausdorff dimension.

There are no Lipschitz space filling curves. For a gentle introduction to where this question leads, see http://eprints.nuim.ie/1626/1/SBuckleycurvesfunctions41.pdf. Theorem 3 is the classical result you want.

Let me add to the examples by Benoit and Victor another Cantor example, this time a straightforward naive one rather than ingenious.

Consider at and just after stage $0$ a closed interval $I$ with the standard (Euclidean) metrics but of length $\frac{11}{10}$. At stage $k>0$ remove the center open interval of length $\frac 1{2^{k-1}\times 11^k}$ of each interval left after the previous stage $k-1$. After all stages $0\ 1\ \ldots$, in the remaining set $C$ in addition to the Euclidean metrics consider also the following pseudo-metrics:

$$d(x\ y) = |x-y| - s_{x\ y}$$

where $s_{x\ y}$ is the sum of the lengths of all removed intervals which are between points $x\ y$. The identity map from Euclidean $C$ to $C$ with the pseudo-metric $d$ is Lipschitz with constant 1. Let $C'$ be the metric space induced by $C$. Then $C'$ is homeomorphic to a nondegenerated closed interval, and the map induced by the identity on $C$ is Lipschitz with constant 1.

Actually, C' is isometric with the unit Euclidean interval $[0;1]$.