Finding a basis for the set of polynomials where f(1)=f(-1)=0

Sorry, but your line of thought is quite wrong.

Your part B is the worst one. Apart from using $x_1$ and $x_2$, which is already bad, you can't prove the set is closed under addition by showing that the sum of two particular element is in the set. You need to show that for any choice of polynomials $p,q\in W$, the sum $p+q\in W$. Similarly for the closure with respect to scalar multiplication.

Let's try a more systematic approach.

Consider the vector space $V$ of polynomials of degree at most $3$ (not *of degree $3$) with real coefficients.

For $a\in\mathbb{R}$, consider $W_a=\{p\in V: p(a)=0\}$. Then $W_a$ is a subspace of $V$. Indeed, the zero polynomial clearly belongs to $W_a$. Moreover, if $p,q\in W_a$, we have $p(a)=0$ and $q(a)=0$; so, if $r(x)=p(x)+q(x)$, we have $$ r(a)=p(a)+q(a)=0+0=0 $$ because of standard laws of polynomials. Finally, if $\lambda\in\mathbb{R}$ and $s(x)=\lambda p(x)$, we have $$ s(a)=\lambda p(a)=0 $$ Therefore $p(x)+q(x)\in W_a$ and $\lambda p(x)\in W_a$.

Your set is $W=W_1\cap W_{-1}$ and is a subspace being the intersection of two subspaces.

In order to find a basis, you can consider the linear map $$ f\colon V\to\mathbb{R}^2,\qquad f(p)=\begin{bmatrix}p(1)\\p(-1)\end{bmatrix} $$ and notice that $W=\ker f$. (Note that linearity of this map is basically the same as the proof above.)

It's easy to find the matrix of $f$ relative to the standard bases on $V$ (that is, $\{1,x,x^2,x^3\}$ and of $\mathbb{R}^2$. It is $$ \begin{bmatrix} 1 & 1 & 1 & 1 \\ 1 & -1 & 1 & -1 \end{bmatrix} $$ A standard Gaussian elimination brings to the reduced row echelon form $$ \begin{bmatrix} 1 & 0 & 1 & 0 \\ 0 & 1 & 0 & 1 \end{bmatrix} $$ and the null space of this matrix has as basis $$ \left\{ \begin{bmatrix}-1\\0\\1\\0\end{bmatrix}, \begin{bmatrix}0\\-1\\0\\1\end{bmatrix} \right\} $$ The polynomials having these coordinates with respect to the standard basis are $x^2-1$ and $x^3-x$, so a basis for $W$ is $$ \{x^2-1,x^3-x\} $$ as was your intuition.