We have $M$ an o-minimal structure. $X \in M^n$ with the induced topology.
I'm reading an article which shows that $X \in M^n$ is definable compact is
equivalent to $X$ being bounded and closed.

Definable compactness of $X$ means that any $M$-definable curve in $X$ is completable.
(a curve in $X$ is a $M$-definable continuous embedding $f: (a,b) \rightarrow X$).
It is said to be completable if $\lim_{x\rightarrow a^{+}}f(x)$ and
$\lim_{x\rightarrow b^{-}}f(x)$ exists.)

When it shows that any definably compact subset $X \in M^n$ is bounded I've got
the feeling that its proof is very complicated. I might be doing something wrong
but I've got the feeling that this proof can be done much more easily.

Here is how it goes: it first shows that definable compactness is preserved under 
projection on the k first coordinates.

Then it proceeds by induction.
let's assume that any definably compact subset $X \in M^n$ is bounded.
Let $X \in M^{n+1}$, then as $p(X)$ is definably compact (by preservation of
definable compactness under projection), it is by induction bounded ($p$ is the
projection onto the $n$ first coordinates).
So the first $n$ coordinates of X are bounded.

Now why can't we also say the projection of X onto his last n coordinates is
definably compact (by preservation of definable compactness under projection),
it is by induction bounded.
So the last n coordinates of X are bounded.

So X is bounded.

(for the proof of "Let $S \in M^n$ be definably compact and let $p : M^n \rightarrow M^k$
be a projection map. Then p(S) is definably compact." is the following:
By induction, it suffices to show that if $S \in M^{n+1}$ is definably
compact then p(S), where $p: M^{n+1} \rightarrow M^n$  denotes projection onto the
first n coordinates, is definably compact as well. For a contradiction, assume
not. Then there is a definable continuous embedding $f: (a, b) \rightarrow p(S)$
such that,say, f does not have a right-hand limit point in $p(S)$. By
o-minimality, for every $a \in p(S)$, the set $S_a = \{b \in M : (a, b) \in S\}$ is the union of
finitely many intervals. Since S is definably compact, $S_a$ is closed and bounded.
Let m(a) be the least element of $S_a$. We now define $g: (a, b) \rightarrow S$ by
$g(x) = (f(x),m(f(x))$. It follows that g does not have a right-hand limit
point in S since f does not have one in $p(S)$, a contradiction.)