No such ideal exists.
A simple Lie algebra $\mathfrak{g}$ over a field $k$ is called absolutely simple if for every algebraic extension $K\vert k$ (or equivalently: for an algebraic closure $K\vert k$) , the scalar extension $K\otimes \mathfrak{g}$ is also simple (note that it necessarily is semisimple; for an example where it is not simple, see below).
One can show that if a Lie algebra over a field $k$ is simple but not absolutely simple, it is the scalar restriction of an absolutely simple Lie algebra over some algebraic extension $K \vert k$. Actually, one can compute $K$ as the (associative) subalgebra of $End_k(\mathfrak{g})$ consisting of those elements that commute with all $ad_\mathfrak{g}(x), x \in \mathfrak{g}$. As far as I know, this was first shown by Jacobson in Duke Math. J., Volume 3, Number 3 (1937), 544-548, doi:10.1215/S0012-7094-37-00343-0, and holds for more general kinds of algebras. I wrote a little overview of that in section 4.1 of my thesis, a lot of it is now reproduced in this answer of mine.
Now to your question: Since scalar extension commutes with direct sums, and ideals of semisimple Lie algebras are direct summands, your question is equivalent to asking whether there exists a simple compact real Lie algebra $\mathfrak{g}$ which is not absolutely simple. But by the above theory, and the fact that the only proper algebraic extension of $\Bbb R$ is $\Bbb C$, the only simple but not absolutely simple real Lie algebras are: the simple complex Lie algebras considered as $\Bbb R$-algebras. The first example maybe being $\mathfrak{sl}_2(\Bbb C)$ viewed as a Lie algebra over $\Bbb R$ (six-dimensional); it is simple, but not absolutely simple, as its scalar extension $\Bbb C \otimes_{\Bbb R} \mathfrak{sl}_2(\Bbb C)$ actually is isomorphic to the sum of two copies of $\mathfrak{sl}_2(\Bbb C)$.
However, none of these scalar restrictions of simple complex Lie algebras correspond to compact Lie groups, e.g. because they obviously contain nilpotent elements.
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 $\mathfrak{gl}_n$.]
Best Answer
From your question, one could assume that there's some contradiction somewhere. Where is it? Yes, $\mathfrak{sl}(2,\mathbb{C})$ is simple and, yes, $\mathfrak{sl}(2,\mathbb{C})$ is semisimple. That's not a problem, since every simple Lie algebra is also semisimple.