Eigenvectors of real symmetric matrices are orthogonal (more discussion) This is an old question, and the proof is here
The proof assumed different eigenvalues with different eigenvectors. 
My question is how about the repeated root? How to guarantee there will not be only one independent eigenvector such that all eigenvectors can form the orthogonal basis of the vector space?
 A: There are really three things going on here:

*

*Eigenvectors corresponding to distinct eigenvalues are all orthogonal.

*A symmetric matrix is diagonalizable whether it has distinct eigenvalues or not. @A.G. proved this just fine already.

*Given a subspace whose dimension is greater than $1$, one can choose a basis of the subspace consisting of orthogonal elements. This is usually proven constructively by applying Gram-Schmidt.

Thus, it is not the case that all pairs of non-parallel eigenvectors of every symmetric matrix are orthogonal to each other. Rather, one can choose an orthogonal basis such that the matrix is diagonal in that basis. Nonetheless, for a symmetric matrix with a repeated eigenvalue, one can also choose a non-orthogonal basis such that the matrix is diagonal in that basis.
A: Assume that for a symmetric matrix $A$ there exists a Jordan block for an eigenvalue $\lambda$ of size more than one, hence there exists at least two  linear independent generalized eigenvectors, i.e. $By=x$ and $Bx=0$ where $B=A-\lambda I$. Estimate $x^TBy$. On one hand it is $0^Ty=0$, on other hand, it is $x^Tx=\|x\|^2$. It gives $x=0$ which is a contradiction with the vectors being linear independent. Hence all chains of generalized eigenvectors are of length one, i.e. they are eigenvectors for $A$.
Addendum: As @Ian correctly noticed, one has to add to the proof that the basis of the corresponding eigen-subspace for $\lambda$ can be chosen orthogonal.
A: An alternative approach to the proof (not using the inner-product method on the question you reference) is to use Schur's Theorem.
Schur's Theorem: Every square matrix $A$ has a factorization of the form $A=QTQ^{\ast}$ where $Q$ is a unitary matrix and $T$ is upper triangular.
Then, if $A$ is symmetric, $T$ must also be symmetric (and hence diagonal).  The columns of $Q$ are the eigenvectors of $A$ (easy to check), $T$ contains the eigenvalues (easy to check), and since $Q$ is unitary, all the columns are orthonormal.
