If $x\in L$ lies in exactly one Cartan subalgebra, then $x$ is regular 
(Question 15.3 in Humphreys's book Introduction to Lie Algebras) Let $L$ be semisimple (char $\mathbb F = 0$), $x\in L$ semisimple. Prove that if $x$ lies in exactly one CSA (Cartan subalgebra), then $x$ is regular.

Definitions. An element $x$ is said regular if $C_L(x) = \{y\in L: [y,x] = 0\}$ is a maximal toral subalgebra. In particular, when $L$ is semisimple, being maximal toral is equivalent to being a CSA which is equivalent to being a minimal Engel subalgebra.
An Engel subalgebra is one of the form $L_0(\mbox{ad }x) = \{y \in L: (\mbox{ad }x)^ky =0 \mbox{ for some integer } k \}, x\in L$.
Remark. This question have been asked here Cartan Subalgebra and regular elements and here $x$ regular $\Leftrightarrow$ $x$ is in exactly one CSA. Both of them lack of an answer. Moreover, the suggestion given in the latter is in my opinion wrong, since taking "$y\in C_L(x)\setminus H$ such that $L_0(\mbox{ad }y)$ is minimal" (I'm assuming he meant minimal among all the $y\in C_L(x)\setminus H$) does not ensure that it is minimal among all Engel subalgebras. I've tried to work a little bit more with this argument as follows. Of course, if $L_0(\mbox{ad }y)$ is not minimal, it will contain a minimal Engel subalgebra, say $L_0(\mbox{ad }z)$ ($L$ has finite dimension). But I can't see how to argue that $x\in L_0(\mbox{ad } z)$ to obtain the absurd of $x$ lying both in $H$ and $L_0(\mbox{ad } z)$.
Some ideas. Suppose $x$ is not regular. On one hand, being semisimple implies that $C_L(x) = L_0(\mbox{ad } x)$. Furthermore, there exists a maximal toral $H\subseteq C_L(x),$ where $H$ is the subalgebra spanned by the semisimple elements of $L$. Since $H = C_L(H)$, it follows that $x\in H$. On the other hand, $x$ being irregular implies that $L_0(\mbox{ad }x) = C_L(x)$ is not miminal, so let us take $L_0(\mbox{ad }y)\subseteq C_L(x)$ properly and minimal. In particular, $y\in C_L(x)$ gives us that $x\in L_0(\mbox{ad }y).$  Both $H$ and $L_0(\mbox{ad }y)$ are CSA, so it remains to prove that they are not the same to obtain a contradiction. $H$ and $L_0(\mbox{ad }y)$ are too generic, I can't see a way to argue that. I'm stuck and out of ideas.
Any insight or help is very much appreciated.
 A: I've finally found a proof which works for any (finite-dimensional) semisimple Lie algebra $L$ over an arbitrary field of characteristic $0$ (and possibly even more). It rests on several "well-known" facts, which, however, when you don't accept them as well-known, are not exactly trivial. In the following, the reference "Bourbaki" is obviously to their books on Lie Groups and Algebras, actually all to chapter VII of that.
Remember that the general definition of a Cartan subalgebra of any Lie algebra is: a subalgebra which is nilpotent and self-normalising. For semisimple Lie algebras, this is equivalent to being a maximal toral subalgebra, where a subalgebra is called "toral" if it is abelian and consists of $ad$-semisimple elements. Cf. 1, 2, 3, 4.
Fact 1: If $x \in L$ is an $ad$-semisimple element, then
$$\lbrace \text{Cartan subalgebras of } L \text{ containing } x \rbrace = \lbrace \text{Cartan subalgebras of } C_L(x) \rbrace. $$
(This is a special case  of Bourbaki, chap. VII, §2 no. 3 Proposition 10, where indeed we don't need $L$ to be semisimple. The inclusion "$\subseteq$" is straightforward in particular in the case of semisimple $L$, from the known equivalence of the definitions in that case. For "$\supseteq$",  let $\mathfrak h$ be a CSA of $C_L(x)$. Clearly $\mathfrak h$ is nilpotent and contains $x$. Further, if $N:=N_L(\mathfrak h)$ is the normaliser of $\mathfrak h$ in $L$, then because $ad(x)(N) \subseteq \mathfrak h$ and $x$ is $ad$-semisimple, there is a decomposition into $ad(x)$-stable vector spaces $N = \mathfrak h \oplus M$ where further $ad(x)(M) = 0$, meaning that $M$ and hence all of $N$ is contained in $C_L(x)$. But then $N = N_{\color{red}{C_L(x)}}(\mathfrak h)$ which, by $\mathfrak h$ being a CSA in $C_L(x)$, is just $=\mathfrak h$.)
Fact 2: If $x \in L$ is a semisimple element, then its centraliser $C_L(x)$ is a reductive Lie algebra.
(See e.g. here, although the proof there uses an algebraically closed field / split semisimple $L$, but then it's really straightforward to reduce to that case: Because the centraliser in a scalar extension is the scalar extension of the centraliser, and a Lie algebra is reductive iff its scalar extension is. Actually, the stronger fact that for any toral subalgebra $\mathfrak a \subset L$, the centraliser $C_L(\mathfrak a)$ is a reductive Lie algebra, is Corollary 3.1.21 in my thesis, where it's also proved via scalar extension. It is also a special case of Bourbaki, chap. VII §1 no. 5 Proposition 13.)
Fact 3: The Cartan subalgebras of a reductive Lie algebra $\mathfrak r$, written as $\mathfrak z \oplus \mathfrak s$ with centre $\mathfrak z$ and semisimple $\mathfrak s$,  are exactly the subalgebras of the form $\mathfrak z \oplus \mathfrak c$ where $\mathfrak c$ is a Cartan subalgebra of $\mathfrak s$.
(This should be clear from the general definition of CSAs as subalgebras which are nilpotent and self-normalising; indeed, in Bourbaki chap VII §2 no. 1 it is mentioned right after that definition that the only CSA of any nilpotent Lie algebra (in particular, an abelian one like $\mathfrak z$) is itself, and it is Proposition 2 that the CSAs of a direct product are the direct products of the CSAs.)
Fact 4: Any semisimple Lie algebra $\mathfrak s \neq 0$ contains more than one CSA.
(Not as trivial as it sounds. But for sure there are many ways to see this in general: Indeed, Bourbaki shows that over any infinite field, every non-nilpotent Lie algebra contains infinitely many CSAs (see my answer to this question). Here is an ad hoc argument: W.l.o.g. $\mathfrak s$ is simple. In a semisimple Lie algebra, every element can be decomposed into a semisimple and a nilpotent part which commute. Basically, we have to find two semisimple elements $x_1, x_2$ which don't commute, because then $x_1$ is contained in a CSA which, by virtue of being abelian (toral!), cannot contain $x_2$, and vice versa, so there are at least two different CSAs. -- Now, if $\mathfrak s$ contains no nonzero nilpotent (this happens, e.g. compact forms over $\mathbb R$), then all its elements are semisimple, but since $\mathfrak s$ is not abelian we are done. So assume there is a nilpotent element $y$. Then, like in Levent's answer, we extend $y$ to an $\mathfrak sl_2$-triple $y,h,z$ and can explicitly find non-commuting semisimple elements from that, e.g. like in his answer, or we note that the inner automorphism $e^{ad(y)}$ of $\mathfrak s$ sends the semisimple $h$ to the semisimple $h \mp 2 y$ (sign depending on how you define $\mathfrak sl_2$-triples), and these two do not commute with each other.)

Now let's go, let $x$ be a semisimple element of $L$. By facts 2, 3, and 4, the Lie algebra $C_L(x)$, in and of itself, contains a unique CSA if and only if it is abelian, if and only if it is itself a CSA (in itself); otherwise it contains several CSAs.
Via fact 1, this translates to:
There is a unique CSA in $L$ containing $x$ if and only if $C_L(x)$ is a CSA in $L$; which by your definition is equivalent to $x$ being regular.
A: Say $x\in\mathfrak{a}$ is the Cartan subalgebra. Note that we can decompose $L=L_0\oplus\bigoplus_{\lambda\in\Omega}L_\lambda$ for some finite set $\Omega\subseteq\mathfrak{a}^*$ that does not contain $0$ ($L_\lambda$ are the eigenspaces of the pairwise commuting operators $ad(h), h\in\mathfrak{a}$). But, $L_0=\mathfrak{a}$ since Cartan subalgebras are self-centralizing (I think this fact is proved in Chapter 8). I claim that $\lambda(x)\neq 0$ for $\lambda\in\Omega$. Indeed, if $\lambda(x)=0$, then $ad(x)$ annihilates $L_\lambda$ so $L_\lambda+\mathbb{F}x$ contains an abelian subalgebra of dimension at least $2$ and contains $x$ (Take an element $y\in L_\lambda$, form $\mathbb{F}x+\mathbb{F}y$). so there exists $0\neq y\in L_\lambda$ such that $ad(x)(y)=\lambda(x)y=0$. Pick $h\in\mathfrak{a}$, $z\in L_{-\lambda}$ such that $y,h,z$ is an $\mathfrak{sl}_2$ triple, i.e. $$
\begin{bmatrix}
0 & 1\\
0 & 0
\end{bmatrix}\mapsto y,\quad\begin{bmatrix}
1 & 0\\
0 & -1
\end{bmatrix}\mapsto h,\quad\begin{bmatrix}
0 & 0\\
1 & 0
\end{bmatrix}\mapsto z
$$ is an isomorphism between $\mathfrak{sl}_2$ and $\mathbb{F}y+\mathbb{F}h+\mathbb{F}z$. Note that $y+z$ is semisimple since its preimage in $\mathfrak{sl}_2$ is semisimple. Moreover, $x$ commutes with $y+z$, so $\mathbb{F}x+\mathbb{F}(y+z)$ is abelian and contains semisimple elements. This contradicts the fact that $x$ is contained in a unique Cartan algebra and we deduce that $\lambda(x)\neq 0$.
This finishes the proof: $L_\lambda$ are also eigenspaces of $ad(x)$ and we have shown that $\lambda(x)$ is never $0$. Thus, the nullspace of $ad(x)$ is exactly $L_0=\mathfrak{a}.$
