Use the fact that a real symmetric matrix is diagonalizable by a real orthogonal matrix.

Proof.

Since $A$ is a real symmetric matrix, it can be diagonalizable by a real orthogonal matrix $P$.
Thus we have
\[P^{-1}AP=D,\]
where $D$ is the diagonal matrix whose diagonal entries are the eigenvalues of $A$.

The matrix $D$ is real since all the eigenvalues are real.
(Note that this is always true for a real symmetric matrix.)

If we have $B^2=A$ for some $B$, then we have
\[(P^{-1}BP)(P^{-1}BP)=P^{-1}AP=D.\]

From this, we see that we can choose $B=PD’P^{-1}$ where $D’$ is the diagonal matrix whose $i$-th diagonal entry is the square root of $i$-th diagonal entries of $D$.
The matrix $D’$ is real since the eigenvalues are non-negative by the assumption, hence the square roots of them are still real.

Since $P$ is also real, the matrix $B$ is real and satisfies
\[B^2=(PD’P^{-1})(PD’P^{-1})=PD’^2P^{-1}=PDP^{-1}=A.\]

Therefore the matrix $B$ satisfies the conditions of the problem.

Related Question.

This problems is a generalization of a liner algebra exam problem of Princeton University.

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