Why is a semisimple Lie-subalgebra of $\mathfrak{sl}(n,\mathbb{C})$ of rank $n-1$ equal to $\mathfrak{sl}(n,\mathbb{C})$? That question came to me, by reading a proof, where the author shows that a semisimple Lie-algebra $\mathfrak{g} \subset \mathfrak{sl}(n,\mathbb{C})$ contains the full subalgebra of diagonal matrices of trace $0$. He does this by showing, that $[E_{1j}, E_{j1}]= E_{11}-E_{jj} \in \mathfrak{g}$ for $j=2,...,n$. He than states, that a semisimple Lie-subalgebra of $\mathfrak{sl}(n,\mathbb{C})$ of rank $n-1$ is equal to $\mathfrak{sl}(n,\mathbb{C})$ which is, what i want to verify.

What i understand as the rank of a finite-dimensional semisimple Lie algebra (over an algebraically closed field of characteristic zero) is the (unique) dimension of a Cartan subalgebra. Equivalently, it is defined to be the dimension of the maximal abelian subalgebra, or in the context of subalgebras of $\mathfrak{sl}(n,\mathbb{C})$, the largest number of (linear combinations of) generators which commute with each other.
I know that for $\mathfrak{sl} (n,\mathbb{C} )$, the Cartan subalgebra may be taken to be the diagonal subalgebra of $\mathfrak{sl} (n,\mathbb{C} )$, consisting of diagonal matrices whose diagonal entries sum to zero. So if this subalgebra has dimension $n-1$ then $\mathfrak{sl} (n,\mathbb{C} )$ has rank $n-1$.

So basically the two questions i have are:
Q1: Why has the Subalgebra of $\mathfrak{g} \subset \mathfrak{sl}(n,\mathbb{C})$ of diagonal trace $0$ matrices, dimension $n-1$? (answered in the comments)
Q2: Why is a semisimple Lie-subalgebra of $\mathfrak{sl}(n,\mathbb{C})$ of rank $n-1$ equal to $\mathfrak{sl}(n,\mathbb{C})$?
 A: To collect a few of the comments together into a coherent picture this boils down to  the classification of maximal semisimple subalgebras of semisimple Lie algebras achieved by Dynkin as Moishe notes.
Let $\mathfrak{g}$ be a simple complex Lie algebra and let $\mathfrak{h}\leq \mathfrak{g}$ be a semisimple subalgebra.
Firstly, we should note that the maximum possible rank of a semisimple subalgebra of $\mathfrak{g}$ is the rank of $\mathfrak{g}$. This follows from the observation that Torsten made which is that any Cartan subalgebra of $\mathfrak{h}$ is contained in a Cartan subalgebra of $\mathfrak{g}$. One way to see this would be to convince your self that semisimple elements of $\mathfrak{h}$ must be semisimple elements of $\mathfrak{g}$.
So a maximal rank subalgebra must have a Cartan subalgebra $\mathfrak{c}$ which is also a Cartan subalgebra of $\mathfrak{g}$. Note that this forces $\mathfrak{h}$ to be the span of $\mathfrak{c}$ together with some of the root spaces. In fact these roots must form a subsystem of our original root system. So the question becomes a combinatorial one: what are the maximal rank root subsystems of the root system of $\mathfrak{g}$?
Well there's a nice way to find these. Draw out the Dynkin diagram of $\mathfrak{g}$ then add a node to turn it into the extended diagram (This node corresponds to the lowest root and if there are two root lengths we can also extend by the lowest short root instead). Then remove any node and the lines attached to it. The remaining diagram is the Dynkin diagram of a maximal rank subalgebra. You can repeat this process recursively to find all the maximal rank subalgebras. If you try this with an $A_n$ root system you should see that you end up with another $A_n$ root system so that the only maximal rank subalgebra of $\mathfrak{sl}_{n+1}$ is itself.
With other types you can get other possibilities though such as $C_i \times C_{n-i} \subset C_n$, $B_4 \subset F_4$ or $A_2\times A_2 \times A_2 \subset E_6$.
You can of course keep removing nodes without extending to find non-maximal subsystems but these won't cover all the non-maximal semisimple subalgebras and we need other techniques.
Here is a excellent answer by Torsten going into this process in more detail for full rank subsystems and here is an answer by me discussing the non-maximal ones. Both contain references to tables of the classification of root subsystems of a root system.
