By inspection you can see that the span $W_2$ of the first and fifth standard basis vectors is an invariant subspace, as is the span $W_3$ of the remaining three standard basis vectors. You therefore have a decomposition $\Bbb R^5=W_2\oplus W_3$ a direct sum of two invariant subspaces (of dimension $2$ and $3$), and you can find the eigenvalues of the restrictions to those subspaces separately, and then combine them.
For $W_2$ you linear operator acts as a (diagonal) reflection interchanging the two coordinates, so you get eigenvalues $1$ and $-1$. For $W_3$ the action is given by the $3\times 3$ sub-matrix at the center, and since you can see that all its rows have the same sum$~2$, it follows that $(1,1,1)$ is an eigenvector with eigenvalue$~2$. Your operator on $W_3$ (still) being given by a symmetric matrix, the orthogonal complement in $W_3$ of the invariant subspace spanned by $(1,1,1)$ will also be an invariant subspace. If you choose any basis of this orthogonal complement (for instance $(1,-1,0),(0,1,-1)$) and compute your operator on its vectors, you find that the operator acts as multiplication by $-1$ on this orthogonal complement. That gives you an eigenvalue $-1$ with multiplicity$~2$ (for $W_3$; remember it already had multiplicity$~1$ for $W_2$, so it will have multiplicity$~3$ overall).