Just want to know if my proof is correct.
First of all, it is easy to check that $f(x)=x^3+x+1$ is irreducible over $\mathbb{F}_2$. This implies that $\frac{\mathbb{F}_2[X]}{(f(x))}$ is indeed a field. In particular, it must be a field extension of $\mathbb{F_2}$.
Let's check that it has exactly 8 elements. Let $\varphi:\mathbb{F}_2[x] \rightarrow \frac{\mathbb{F}_2[x]}{(f(x))}$ be the canonical projection and $\alpha = \varphi(x)$. Then $f(x)$ is the minimal polynomial of $\alpha$ over $\mathbb{F}_2$, since it is monic and irreducible. Then $\frac{\mathbb{F}_2[X]}{(f(x))}$ is a 3-dimensional vector space over $\mathbb{F}_2$ and the set $\{ 1, \alpha, \alpha^2 \}$ is a basis. Since $\mathbb{F}_2$ has only 2 elements, there are only 8 possible linear combinations.
I think your verification that $\Bbb F_2[X]/(x^3+x+1)$ contains $8$ elements is overly complicated. For every nonzero polynomial $P\in K[X]$, the quotient $K[X]/P$ is a $K$-vector space of dimension $\deg P$ (with as possible basis the classes of the monomials $X^k$ for $0\leq k<\deg P$). A vector space over $\Bbb F_2$ of dimension $3$ has $2^3=8$ elements. (Of course you do need $P$ to be irreducible for the quotient to be a field, but for the dimension as $K$-vector space only the degree of the polynomial matters.)