totally antisymmetric structure constants and metric of a Lie Algebra Given a (semisimple) Lie Algebra, and choosing a base $\{T^a\}$, the structure constants $f^{ab}_c$ are defined by
$
[ T^a, T^b ] = f^{ab}_cT^c
$
The metric tensor is matrix obtained from the Killing form between two elements of the base
$
g^{ab}\equiv K(T^a, T^b) = f^{ai}_jf^{bj}_i
$
I am interested in the relation between these statements about the chosen base.
(a) The structure constants $f^{ab}_c$ are totally antisymmetric in the indexes $a, b, c$. In particular, $f^{ab}_c = -f^{cb}_a$
(b) The quantity $\sum_a T^aT^a$ commutes with the algebra, that is, it is a Casimir operator
$
[ \sum_a T^aT^a, T^b] = 0 \quad \forall\; b
$
(c) The metric tensor is proportional to the identity $g^{ab}\propto \delta^{ab}$, that is, for the chosen base the Killing form is $K(T^a,T^b) \propto \delta^{ab}$.
Edit
To be clear, I am not asking if a), b), or c) always hold, but rather, say, does a) imply c)? Or viceversa? For instance, c) implies b) as also discussed in the answer below. Also, a) implies b), as this follows from straighforward computation of the commutator $[T^aT^a,T^b]$.
 A: I am not comfortable with index notation so I'll translate everything into language that's more familiar to me, especially because it will clarify the sense in which I don't think your question c) makes sense.
A finite-dimensional Lie algebra $\mathfrak{g}$ has a Killing form
$$K(X, Y) = \text{tr}(\text{ad}_X \text{ad}_Y)$$
where $\text{ad}_X$ refers to the adjoint action of $[X, -]$ on $\mathfrak{g}$. This is your "metric tensor" (I would not use this term for the Killing form because in general it is neither guaranteed to be nondegenerate nor definite). By Cartan's criterion, over a field of characteristic zero the Killing form is nondegenerate iff $\mathfrak{g}$ is semisimple. From here on the ground field will always have characteristic zero.
When the Killing form is nondegenerate, we can define a canonical element of the universal enveloping algebra $U(\mathfrak{g})$ called the Casimir element by picking a basis $X_i$ of $\mathfrak{g}$, calculating its dual basis $Y_i$ with respect to the Killing form (so that $K(X_i, Y_i) = \delta_{ij}$), and computing $\sum X_i Y_i \in U(\mathfrak{g})$. The Casimir element always lies in the center of the universal enveloping algebra, which is to say that it always commutes with the elements of $\mathfrak{g}$; this addresses your question b).
The Killing form and the Lie bracket together allow us to define a trilinear form
$$K([X, Y], Z) = \text{tr}([\text{ad}_X, \text{ad}_Y] \text{ad}_Z)$$
which I believe is called the Cartan $3$-form. This trilinear form is always (totally) antisymmetric. It is clearly antisymmetric in the first two variables so it suffices to show that $K([X, Y], Z) = -K([X, Z], Y)$. To do this we just compute that
$$K([X, Y], Z) = \text{tr}(\text{ad}_X \text{ad}_Y \text{ad}_Z) - \text{tr}(\text{ad}_Y \text{ad}_X \text{ad}_Z)$$
whereas
$$K([X, Z], Y) = \text{tr}(\text{ad}_X \text{ad}_Z \text{ad}_Y) - \text{tr}(\text{ad}_Z \text{ad}_X \text{ad}_Y) = \text{tr}(\text{ad}_Y \text{ad}_X \text{ad}_Z) - \text{tr}(\text{ad}_X \text{ad}_Y \text{ad}_Z)$$
where we use the cyclicity of the trace. Alternatively we can write $K([X, Z], Y) = K(Y, [X, Z])$ and $K([X, Y], Z) = - K([Y, X], Z)$ so the desired condition is equivalent to the condition that the Killing form is invariant. I believe this is equivalent to your question a) (in the semisimple case), since in order to ask whether the structure constants are totally antisymmetric you need to raise an index and that should correspond to using the Killing form in this way.
Your question c) does not make sense to me as written, and this is why I have been careful to use more ordinary mathematical language. There is no such thing as "the identity tensor $\delta^{ab}$" as a basis-independent object; the identity is a tensor $I^a_b$. Given a fixed basis $X_i$ you can ask whether the Killing form satisfies $K(X_i, X_j) \propto \delta_{ij}$, and you can also ask whether there exists such a basis; is that what you meant to ask? If so it's confusing because you chose the basis in advance.
Over $\mathbb{C}$ such a basis always exists, while over $\mathbb{R}$ it will not exist unless $\mathfrak{g}$ is compact, because this condition requires the Killing form to be negative definite (it is never positive definite) and it is indefinite in the noncompact case, e.g. for $\mathfrak{g} = \mathfrak{sl}_2(\mathbb{R})$.
So, to summarize:

a) holds, suitably interpreted, in any semisimple Lie algebra.


b) holds in any semisimple Lie algebra.


c) holds, suitably interpreted, in any semisimple Lie algebra over $\mathbb{C}$ but over $\mathbb{R}$ only holds for compact semisimple Lie algebras.

