If $\langle Ta, a\rangle \in \mathbb{R}$ for all $a$ then $T$ is self-adjoint

Here are two hints. You want to use the following two things:

1. $\langle T \alpha, \alpha \rangle = \langle \alpha, T^{\ast} \alpha \rangle$ (by the definition of the adjoint).
2. $\langle T \alpha, \alpha \rangle = \overline{ \langle \alpha, T \alpha \rangle}$.

The tricky part is then proving the following claim:

Let $A$ be an operator on a complex inner product space. Then $A = 0$ if and only if $\langle Ax, x \rangle = 0$ for all $x$.

To prove this, you need to prove a variant of the polarization identity, which should be

$$ \langle Ax, y \rangle = \frac{1}{4} \left( \langle A(x + y), x + y \rangle - \langle A(x - y), x -y \rangle + i \langle A(x + iy), x + iy \rangle - i \langle A(x - iy), x - iy \rangle \right) $$

Apply this claim to $A = T - T^{\ast}$.

Edit: Alternatively, you can use the fact that $T - T^{\ast}$ is skew-skymmetric; that is, its adjoint is equal to its negative. Then it is a normal operator, and see if you can apply the spectral theorem.

First note that $$ \langle T(a), a \rangle = \overline{\langle a, T(a) \rangle} = \langle a, T(a) \rangle = \langle T^*(a), a \rangle $$ for any $a \in V$. We get the second equality since the conjugate of a real number is itself and the other equalities are obtained from definitions. We can use the derived equality, $\langle T(a), a \rangle = \langle T^*(a), a \rangle$, for $a = x + y$ and $a = x + iy$ for any $x, y \in V$. Using properties of linear transformations and inner products, we get $$ \langle T(x + y), x + y \rangle = \langle T^*(x + y), x + y \rangle \\ \langle T(x) + T(y), x + y \rangle = \langle T^*(x) + T^*(y), x + y \rangle \\ \langle T(x), x \rangle + \langle T(x), y \rangle + \langle T(y), x \rangle + \langle T(y), y \rangle = \langle T^*(x), x \rangle + \langle T^*(x), y \rangle + \langle T^*(y), x \rangle + \langle T^*(y), y \rangle. $$ But $\langle T(x), x \rangle = \langle T^*(x), x \rangle$ and $\langle T(y), y \rangle = \langle T^*(y), y \rangle$, so those terms cancel to give $$ \langle T(x), y \rangle + \langle T(y), x \rangle = \langle T^*(x), y \rangle + \langle T^*(y), x \rangle.\tag{1} $$ Similarly, for $a = x + iy$, you just need the extra step of taking out the factor $i$ and you will get $$ -i\langle T(x), y \rangle + i\langle T(y), x \rangle = -i\langle T^*(x), y \rangle + i\langle T^*(y), x \rangle. $$ Dividing out by $-i$ gives $$ \langle T(x), y \rangle - \langle T(y), x \rangle = \langle T^*(x), y \rangle - \langle T^*(y), x \rangle. \tag{2} $$ Add equations $(1)$ and $(2)$ and divide by $2$ to get $$ \langle T(x), y \rangle = \langle T^*(x), y \rangle \\ \langle (T - T^*)(x), y \rangle = 0 $$ for any $x, y \in V$. So we can take $y = (T - T^*)(x)$ to get $\langle (T - T^*)(x), (T - T^*)(x) \rangle = 0$ which implies $(T - T^*)(x) = 0$ for any $x \in V$. Thus, $T - T^* = O$, ie. $T = T^*$ which means $T$ is self adjoint.

I don't think this is true for real inner product space since we can't use $x + iy$ so we can't prove this.