On a field $K$ with $char(K)$ not equal to 2, all irreducible polynomials of a quadratic extension are separable. The proof is straightforward: Assume the opposite, namely $P=X^2+aX+b = (X-\alpha)(X-\beta)$ with $P \in K[X]$, so $a = -\alpha -\beta$. Assuming $\alpha$ and $\beta$ are the same, we get: $a = -2\alpha$. We can solve for $\alpha$ and see that $\alpha \in K$.
With $char(K)=2$, we cannot do that. I figured out that $Q=X^2+X+1$ is irreducible in $\mathbb{F}_2$ and separable because $Q$ and $Q'$ are coprime. How do the roots look like in a splitting field of $Q$?
Let $F$ be a field, write $F[x]$ for the ring of polynomials with coefficients in $F$, and let $f$ be in $F[x]$ and irreducible over $F$. Then $(f(x))$ (which means the ideal generated by $f(x)$, which is just the set of all multiples of $f(x)$ in $F[x]$) is a maximal ideal in $F[x]$, so the quotient ring, $K=F[x]/(f(x))$, is a field (and an extension field of $F$ --- see note below). The elements of $K$ are cosets of the ideal $(f(x))$ in $F[x]$, so they are of the form $g(x)+(f(x))$ for various $g(x)$ in $F[x]$. The element $x+(f(x))$ is a zero of the polynomial $f(x)$ in $K$.
How this works in the particular case of $f(x)=x^2+x+1$ over the field of two elements is elaborated in the comments on the question.
Readers not familiar with maximal ideals, quotient rings, and other concepts used above are referred to any textbook that covers Field Theory.
Note: we identify the set of all elements $c+(f(x))$ of $K$ with $c$ in $F$ as the inclusion of $F$ in $K$, justifying the view of $K$ as an extension of $F$.