Weyl's theorem states
Let $\mathfrak{g}$ be a finite dimensional semisimple Lie-algebra over an algebraically closed field with characteristic 0. Then all finite dimensional representations are semisimple.
One can also look at the other direction
Let $\mathfrak{g}$ be a finite dimensional Lie-algebra over an algebraically closed field with characteristic 0. If all finite dimensional representations are semisimple then $\mathfrak{g}$ is semisimple.
I want to prove that direction. For that I considered the adjoint representation giving us a direct sum of irreducible subrepresentations. These correspond to the ideals of $\mathfrak{g}$. Now, I would like to show that these ideals are simple (or semisimple). But now, I am stuck showing this. I have tried to look at abelian ideals of these ideals and tried to show that they are 0, but it has not worked. Any hints would be appreciated.
Update: I have expanded this answer to try and provide more details in response to the comment below.
Let $\mathfrak g$ be a finite-dimensional Lie algebra over an algebraically closed field of characteristic zero, and let $\mathrm{Rep}_{fd}(\mathfrak g)$ denote the category of finite-dimensional representations of $\mathfrak g$.
Claim: If $\mathrm{Rep}_{fd}(\mathfrak g)$ is semisimple (that is every object of $\mathrm{Rep}_{fd}(\mathfrak g)$ is semisimple) then $\mathfrak g$ is a semisimple Lie algebra.
Definition: Let $\mathfrak g$ be a Lie algebra over a field of characteristic zero. We say that $\mathfrak g$ is reductive if its radical (i.e. its largest solvable ideal) is equal to its centre. If the radical $\mathrm{rad}(\mathfrak g)$ is equal to the centre $\mathfrak z(\mathfrak g)$, then the Levi decomposition of $\mathfrak g$ will split as a direct sum of a semisimple Lie algebra and the centre of $\mathfrak g$.
Claim 1: If the adjoint representation $(\mathfrak g, \mathrm{ad})$ is completely reducible, then $\mathfrak g$ is reductive.
If $(\mathfrak g, \mathrm{ad})$ is completely reducible, we may write $\mathfrak g$ as a direct sum of irreducible subrepresentations,. Suppose this decomposition takes the form $$ \mathfrak g = \bigoplus_{i=1}^k \mathfrak p_i \oplus \mathfrak g^{\mathfrak g}, $$ where for a $\mathfrak g$-representation $(V,\rho)$, we write $V^{\mathfrak g} = \{v \in V: \rho(g)(v)=0, \forall g\in \mathfrak g\}$, hence when $V$ is semisimple, $V^{\mathfrak g}$ is the isotypical summand of $V$ corresponding to the trivial representation. Now each $\mathfrak p_i$ is an irreducible $\mathfrak g$-representation, and hence a minimal ideal in $\mathfrak g$, thus each $\mathfrak p_i$ is a Lie algebra which has no proper ideals.
Now because the ideals $\mathfrak p_i$ form a direct sum, $[\mathfrak p_i,\mathfrak p_j]\subseteq \mathfrak p_i \cap \mathfrak p_j = \{0\}$. Thus as the $\mathfrak p_i$ are, by definition, not contained in $\mathfrak g^{\mathfrak g}$ we must have $[\mathfrak p_i,\mathfrak p_i]\neq 0$ and hence, by minimality of $\mathfrak p_i$, it follows that $[\mathfrak p_i,\mathfrak p_i]= \mathfrak p_i$.
Thus we see that $\mathfrak g = \mathfrak s \oplus \mathfrak g^{\mathfrak g}$, where $\mathfrak s$ is a direct sum of non-abelian simple Lie algebras (and hence is semisimple) and $\mathfrak g^{\mathfrak g}$ is an abelian ideal on which $\mathfrak g$ acts by zero, i.e. it is the centre of $\mathfrak g$, and the claim is established.
Claim 2: If $\mathfrak g$ is any Lie algebra which has $\mathfrak{gl}_1$ as a quotient, then $\mathrm{Rep}_{fd}(\mathfrak g)$ is not semisimple.
Proof: The point here is that the quotient map $q\colon \mathfrak g \to \mathfrak{gl}_1$ can be used to pull back any representation of $\mathfrak{gl}_1$ to become a representation of $\mathfrak{g}$. In particular, if $\mathfrak{gl}_1$ has any representations which are not semisimple then so will $\mathfrak g$. But the map $t\mapsto t\left(\begin{array}{cc}1 & 1\\ 0 & 1\end{array}\right)$ gives a representation of $\mathfrak{gl}_1$ on $\mathsf k^2$ which has a unique proper submodule (the span of $(1,0)^\intercal$) and hence cannot be semisimple.
Finally, to establish the original Claim, notice that if $\mathrm{Rep}(\mathfrak g)$ is semisimple then certainly $(\mathfrak g, \mathrm{ad})$ is a semisimple representation and so by Claim 1, $\mathfrak g$ is reductive. But then if $\mathfrak z(\mathfrak g)\neq 0$, then since $\mathfrak g = \mathfrak s \oplus \mathfrak z(\mathfrak g)$ where $\mathfrak s$ is a semisimple ideal, it follows that $\mathfrak g/\mathfrak s \cong \mathfrak z(\mathfrak g)$, and as $\mathfrak z(\mathfrak g)$ is abelian, if $\lambda \in \mathfrak z(\mathfrak g)^*$ is a nonzero linear functional on $\mathfrak z(\mathfrak g)$, then $\lambda \colon \mathfrak z(\mathfrak g) \to \mathfrak{gl}_1$ is a surjective homomorphism of Lie algebras. Since $q\colon \mathfrak g \to \mathfrak g/\mathfrak s \cong \mathfrak z(\mathfrak g)$ is also surjective, it follows that $\lambda \circ q$ is a surjective homomorphism from $\mathfrak g$ to $\mathfrak{gl}_1$.
Therefore, it follows from Claim $2$ that if $\mathrm{Rep}_{fd}(\mathfrak g)$ is semisimple we must have $\mathfrak z(\mathfrak g)=\{0\}$ and $\mathfrak g$ is semisimple as required.
[Note that an abelian Lie algebra $\mathfrak a$ is just a $\mathsf k$-vector space equipped with the zero Lie bracket. Thus $\mathfrak a$ can be written as a direct sum of one-dimensional subspaces which are also subalgebras. The only one-dimensional Lie algebra is the trivial one because the alternating property of the bracket forces it to vanish, and this Lie algebra is usually denoted $\mathfrak{gl}_n$.]