The proof of this property is not so easy as those of the basic properties of eigenvalues and eigenvectors. It can be shown by induction, or by explicit construction (see eg here)
I like to visualize the property in this way:
We know that an hermitian matrix with $n$ distict eigenvalues has $n$ eigenvectors that are not only LI (as in general matrices) but, more than that, orthogonal. We also know that this matrix is diagonalizable, with unitary $U$ (both properties are easy to prove).
Now, if our hermitian matrix happens to have repeated (degenerate) eigenvalues, we can regard it as a perturbation of some another hermitian matrix with distinct eigenvalues.
By a continuity argument, we should see that the matrix perturbation than transforms different (but perhaps close) eigenvalues into coincident ones, cannot make the orthogonal eigenvectors linearly dependent.
Put in other way: an hermitian matrix $A$ with repeated eigenvalues can be expressed as the limit of a sequence of hermitian matrices with distinct eigenvalues. Because all members of the sequence have $n$ orthogonal eigenvectors, by a continuity argument, they cannot end in LD eigenvectors.
This approach leads to a nice intuition, IMO, and it can be formalized. But for a formal proof the other methods are to be preferred.
By contradiction:
If $v_1 + v_2$ is an eigenvector of A then there exists and eigenvalue $\lambda$ so that $$ A(v_1 + v_2) = \lambda(v_1 + v_2) = \lambda v_1 + \lambda v_2.$$
However since $v_1$ and $v_2$ are eigenvectors and $A$ is linear we have
$$ A(v_1 + v_2) = A(v_1) + A(v_2) = \lambda_1 v_1 + \lambda_2v_2.$$
Therefore
$$ \lambda v_1 + \lambda v_2 = \lambda_1 v_1 + \lambda_2v_2$$
$$ \iff$$
$$ (\lambda - \lambda_1) v_1 + (\lambda - \lambda_2)v_2 = 0. $$
Since $\lambda_1 \neq \lambda_2$, $v_1$ and $v_2$ are linearly independent so
$$ \lambda - \lambda_1 = 0 \qquad \lambda-\lambda_2 = 0.$$
So $ \lambda = \lambda_1 = \lambda_2 $ which is a contradiction.
Best Answer
Expansion of my comment above:
If $B$ is a positive semidefinite (symmetric) matrix then there exist an orthonormal matrix $S$ (satisfying $S'S=I$) and a diagonal matrix $\Lambda$ with nonnegative diagonal entries ($\Lambda=\text{diag}(\lambda_1,\ldots,\lambda_n)$ where each $\lambda_j\geq 0$) such that $$ B=S'\Lambda S. $$ Then $\sqrt{B}$ is usually defined as $S'\sqrt{\Lambda}S$ where $\sqrt{\Lambda}=\text{diag}(\sqrt{\lambda_1},\ldots,\sqrt{\lambda_n})$. You can now check via direct multiplication that $$ \sqrt{B}\sqrt{B}=S'\sqrt{\Lambda}(SS')\sqrt{\Lambda}S=S'\sqrt{\Lambda}\sqrt{\Lambda}S=S'\Lambda S=B. $$ By the way, the eigenvalues of $\sqrt{B}$ are the square roots of the eigenvalues for $B$.