If $\mathfrak{g}$ is a complex, semisimple, finite-dimensional Lie algebra and $\mathfrak{h} \subseteq \mathbb{g}$ a Cartan subalgebra, then $\mathfrak{h}$ is abelian and self-centralizing, i.e. $Z_\mathfrak{g}(\mathfrak{h}) = \mathfrak{h}$ (this is, for example, show in Humphrey’s book).
If $\mathfrak{h}$ was not a maximal abelian subalgebra with respect to the inclusion, then $\mathfrak{h}$ would be properly contained is some maximal abelian subalgebra $\mathfrak{a}$ of $\mathfrak{g}$. But then $\mathfrak{a} \subseteq Z_\mathfrak{g}(\mathfrak{h})$, contradicting $\mathfrak{h}$ being self-centralizing. So if $\mathfrak{g}$ is finite-dimensional it is true that any Cartan subalgebra is a maximal abelian subalgebra. (I don’t know what happens in the infinite-dimensional case.)
The example in Wikipedia, namely
$$
\mathfrak{a} =
\left\{
\begin{pmatrix}
0 & A \\
0 & 0
\end{pmatrix}
\in \mathfrak{sl}_{2n}(\mathbb{C})
\,\middle|\,
A \in \mathfrak{gl}_n(\mathbb{C})
\right\},
$$
shows that while Cartan-subalgebras are maximal abelian subalgebras with respect to the inclusion, they are not necessarily of maximal dimension among all abelian subalgebras: Every Cartan subalgebra of $\mathfrak{sl}_{2n}(\mathbb{C})$ has dimension $2n-1$ (for example the traceless diagonal matrices), but $\mathfrak{a}$ has dimension $n^2$.
What this tells us that $\mathfrak{a}$ does not contain a Cartan subalgebra of $\mathfrak{sl}_{2n}(\mathbb{C})$ (strictly speaking we only get this for $n > 1$, but for $n = 1$ this is also easy to see).