$x$ regular $\Leftrightarrow$ $x$ is in exactly one CSA

Here's a statement and a proof given in a Lie Algebra course (in the tutorial):

Let $L$ be a semisimple Lie algebra over a field $F$ with $\text{char} F=0$. Let $x\in L$ be a semisimple element. Prove that $x$ is regular if and only $x$ is an element of exactly one CSA.

Proof (this is just one direction: "not regular $\Rightarrow$ belongs to more than one CSA". The other direction is clear)

Suppose $x$ is not regular. So, $C_L(x)$ is not a maximal toral subalgebra of $L$. Using the adjoint representation, we can assume that $L$ is a matrix algebra with $x$ a diagonal matrix. $C_L(x)$ properly contains a maximal toral subalgebra $H$ with $x \in H$.

Now, $C_L(x)$ is not abelian and therefore we can take some $y \in C_L(x) \setminus H$ with $L_0(\text{ad}y)$ minimal. This $L_0(\text{ady})$ is minimal Engel and therefore a CSA. It is not $H$ because $y \in L_0(\text{ad}y)\setminus H$. So $x$ is an element of two different CSAs.

My confusion:

1. I don't understand the reasoning in "Now, $C_L(x)$ is not abelian and therefore we can take some $y \in C_L(x) \setminus H$ with $L_0(\text{ad}y)$ minimal". The problem is with the therefore part.
2. I don't see where viewing $L$ as a matrix algebra helps in the proof given.
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