Metric: $X$ is compact if $X$ is finite i am having again some maths problems i don't understand.
Unfortunately, in this case, i don't have even an idea how to start with :-/
This is my Problem:
Let $X$ a non-empty set. We define a metric $d$ on $X$ with $$d(x,y)=\begin{cases}1 & x \neq y\\0 & x = y\end{cases}$$
How can i show that $X$ is compact if $X$ is finite?
Is this maybe possible with example of sets? If i am not totally wrong the metric should be a discrete metric. Right? I have no idea...
 A: Let $\{U_{\alpha}\}$ be any cover of $X$. Let $x_1 \in X$. Then there exists some $U_{\alpha}$ that contains $x_1$ by definition. Continue in this way for all $x_{i} \in X$. You will get a family of $U_{\alpha_i}$ that is finite by assumption.
This procedure has nothing to do with the metric, and that is because every topology on a finite set makes it compact.

Using your definition from the comments: If you have only finitely many elements in $X$,, while  the sequence has infinitely many elements of $X$, one among them must occur infinitely many times. What can you construct by knowing this?
A: Since you're using a definition of compactness in terms of sequences (rather than open covers), I suggest showing that if all the terms in a sequence are in a finite set $X$, then there is a subsequence all of whose terms are the same, i.e., a constant sequence.  Then show (if you don't already have this result) that every constant sequence in a metric space converges.  
A: Since $X$ is finite, $X = \{x_1,\ldots,x_n\}$ say, an infinite sequence of elements of $X$ can't have only finitely many of each of $x_1,\ldots,x_n$.
It follow that at least one of $x_1,\ldots,x_n$, must occur infinitely often in the sequence.
Thus the sequence has an infinite subsequence which is constant, hence converges to that constant.

Therefore (using your definition), $X$ is compact.
