I think that you forgot the following fact:
Suppose that $A$ is an $n \times n$ real symmetric matrix. Suppose that $Au = \lambda u$ and $Av = \mu v$, in which $u$, $v \in \mathbb{R}^{n}$ and $\lambda$, $\mu$ are distinct real numbers. Then $u^{\mathrm{T}} v = 0$.
There is another fact:
Suppose that $A$ is an $n \times n$ matrix. Suppose that $k$ is a number. Suppose that $u_1$, $u_2$, $\dots$, $u_m$ are $n \times 1$ matrices such that
$$
Au_i = \color{red}{k} u_i.
$$
Suppose that $c_1$, $c_2$, $\dots$, $c_m$ are numbers. If $u = c_1 u_1 + c_2 u_2 + \dots + c_m u_m$ is nonzero, then $Au = \color{red}{k} u$, which means that $u$ is an eigenvector of $A$.
Okay. Take any $n \times n$ real symmetric matrix $A$. Suppose that $\lambda$ is an eigenvalue of $A$. Suppose that you have found all $m$ real linearly independent solutions to $AX = \lambda X$:
$$
u_1, u_2, \dots, u_m.
$$
By Gram-Schmidt, there exists real numbers $c_{i,j}$ such that
$$
\begin{aligned}
& v_1 = c_{1,1} u_1, \\
& v_2 = c_{1,2} u_1 + c_{2,2} u_2, \\
& \cdots \cdots \cdots \cdots, \\
& v_m = c_{1,m} u_1 + c_{2,m} u_2 + \dots + c_{m,m} u_m,
\end{aligned}
$$
and that
$$
v_i^{\mathrm{T}} v_j = \begin{cases}
1, & i = j; \\
0, & i \neq j.
\end{cases}
$$
By the second fact that I have listed above, $v_1$, $v_2$, $\dots$, $v_m$ are still eigenvectors of $A$.
Suppose that $\mu$ is another eigenvalue of $A$. Suppose that you have found $p$ real orthonormal solutions to $AX = \mu X$:
$$
w_1, w_2, \dots, w_p.
$$
By the first fact that I have listed above, $w_i^{\mathrm{T}}$ must be orthogonal to $v_j$; you do not have to apply Gram-Schmidt to the list
$$
v_1, v_2, \dots, v_m, w_1, w_2, \dots, w_p,
$$
since they are already orthonormal.
I hope that what I have said is helpful.
