If $A^2 = I$, then $A$ is diagonalizable, and is $I$ if $1$ is its only eigenvalue

Let $A$ be a square matrix of order $n$ such that $A^2 = I$.

1. Prove that if $1$ is the only eigenvalue of $A$, then $A = I$.
2. Prove that $A$ is diagonalizable.

For (1), I know that there are two eigenvalues which are $1$ and $-1$, how do I go about proving what the question asks me?

If $A \neq I$, then there is a (non-zero) vector $v$ such that $v \neq Av$, which is to say $v-Av \neq 0$. But then we have $$ A(v-Av) = Av - A^2v = Av - Iv = Av - v = -(v-Av) $$ which means that $v-Av$ is a non-zero eigenvector of $A$ with eigenvalue $-1$. Thus we have proven $$ A \neq I \implies A\text{ has } -1\text{ as an eigenvalue} $$ which is the contraposition of what we were asked to prove, and we are therefore done.

You are right, eigenvalues are $1$ and $-1$. For diagonalizable part: $$A^2=I$$ Therefore the characteristic polynomial is $( \lambda-1)(\lambda+1)=0$, hence characteristic polynomial splits. So $A$ is diagonalizable.

Hint for 2. Assume "optimistically" that every vector $v$ decomposes as a sum of eigenvectors of $A$: $$ v = v_{1} + v_{-1}. $$ Apply $A$ to both sides to get a system of two equations in two unknowns, and solve formally for $v_{1}$ and $v_{-1}$.

Armed with the expressions from the preceding paragraph, show that you have actually found eigenvectors of $A$, and therefore have proven that $\mathbf{R}^{n}$ (or whatever field of scalars, so long as the characteristic isn't $2$) decomposes as a sum of eigenspaces of $A$.