As discussed in the comments, the answer is no, for a striking reason: In general, $\mathfrak{a} + \overline{\mathfrak{a}}$ is not even a subalgebra.

As example, take $\mathfrak{g} = \mathfrak{su}_2$ which we can e.g. (following Wikipedia) define as the three-dimensional real Lie algebra with basis $u_1, u_2, u_3$ satisfying

$$\left[u_3, u_1\right] = 2u_2, \quad \left[u_1, u_2\right] = 2u_3, \quad \left[u_2, u_3\right] = 2u_1.$$

It is well-known that the complexification $\Bbb C\otimes\mathfrak{g}$ is isomorphic (as complex Lie algebra) to $\mathfrak{sl}_2(\Bbb C)$. Concretely, an isomorphism is given by
$$f(-\frac{i}{2}\otimes u_1 -\frac{1}{2}\otimes u_2) = \pmatrix{0&1\\0&0},$$
$$ f(-\frac{i}{2}\otimes u_1 +\frac{1}{2}\otimes u_2)=\pmatrix{0&0\\1&0},$$
$$ f(-i\otimes u_3) = \pmatrix{1&0\\0&-1}$$

One sees from this that $f$ identifies the conjugation action $a\otimes g \mapsto \bar a\otimes g$ on $\Bbb C\otimes \mathfrak{g}$ with
$$\pmatrix{a&b\\c&-a} \mapsto \pmatrix{-\bar a&-\bar c\\-\bar b&\bar a}$$
on $\mathfrak{sl}_2(\Bbb C)$ (and $\mathfrak{g}$ with the fixed point set of that, the traceless skew-hermitian matrices, in particular $u_1 = \begin{pmatrix}
0 & i \\
i & 0
\end{pmatrix}, \quad
u_2 = \begin{pmatrix}
0 & -1 \\
1 & 0
\end{pmatrix}, \quad
u_3 = \begin{pmatrix}
i & 0 \\
0 & -i
\end{pmatrix}$ ).

But then if we choose $\mathfrak{h} = \mathfrak{g}_{\Bbb C}$, and $\mathfrak{a}$ as the one-dimensional algebra generated by $\pmatrix{0&1\\0&0}$ (which is maximal abelian in $\mathfrak{h}$), we have $\overline{\mathfrak{a}} = \pmatrix{0&0\\*&0}$, hence $$\mathfrak{a} + \overline{\mathfrak{a}} = \lbrace \pmatrix{0&b\\c&0}: b,c\in \Bbb C\rbrace$$
which is not closed under the Lie bracket, hence no Lie subalgebra of $\mathfrak{g}_{\Bbb C}$.

**NB.** If (in the special case $\mathfrak{h} = \mathfrak{g}_{\Bbb C}$) $\mathfrak{a}$ is chosen as a *Cartan* subalgebra of $\mathfrak{g}_{\Bbb C}$, in some low-dimensional examples it seems to me that $\overline{\mathfrak{a}} = \mathfrak{a}$ and hence the assertion is true. Maybe one can go from there to a case of more general $\mathfrak{h}$. However, I am not even sure if/why this should hold in the special case. (I've asked this as a new question MSE/2756715: Are all CSA's in $\mathfrak{g}_{\Bbb C}$ invariant under the conjugation action?)

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