No. Take an elliptic curve $X$, and let $p \in X$ be a $3$-torsion point. Then there is a plane model of $X$ as a cubic curve in $\mathbb{P}^2(\mathbb{C})$, such that $p$ is an inflexion point.

If $L$ is the corresponding inflexion line, then $U=X-L = X-\{p\}$ is dense in $X$ and affine (being the complement of a hyperplane section).

Now $X$ and $U$ are not homeomorphic, for instance because their fundamental groups are not isomorphic; in fact $$\pi_1(X) = \mathbb{Z} \times \mathbb{Z}, \quad \pi_1(U) = \mathbb{Z} \ast \mathbb{Z}.$$

Let $k$ be an algebraically closed field, ad take $X=\mathbb{P}^2_k$, $U=\mathbb{A}^2_k$.

Then $X$ and $U$ are not homeomorphic, since $U$ contains two disjoint, Zariski-closed, irreducible subsets made of more than one point (think of two parallel lines), but this is not possible in $X$ because of Bézout theorem.

No. Take an elliptic curve $X$, and let $p$ be a $3$-torsion point. Then there is a plane model of $X$ as a cubic curve in $\mathbb{P}^2(\mathbb{C})$, such that $p$ is an inflexion point.

If $L$ is the corresponding inflexion line, then $U=X-L = X-\{p\}$ is dense in $X$ and affine (being the complement of a hyperplane section).

Now $X$ and $U$ are non-homeomorphic, for instance because they have non-isomorphic fundamental group; in fact $$\pi_1(X) = \mathbb{Z} \times \mathbb{Z}, \quad \pi_1(U) = \mathbb{Z} \ast \mathbb{Z}.$$

No. Take an elliptic curve $X$, and let $p \in X$ be a $3$-torsion point. Then there is a plane model of $X$ as a cubic curve in $\mathbb{P}^2(\mathbb{C})$, such that $p$ is an inflexion point.

If $L$ is the corresponding inflexion line, then $U=X-L = X-\{p\}$ is dense in $X$ and affine (being the complement of a hyperplane section).

Now $X$ and $U$ are not homeomorphic, for instance because their fundamental groups are not isomorphic; in fact $$\pi_1(X) = \mathbb{Z} \times \mathbb{Z}, \quad \pi_1(U) = \mathbb{Z} \ast \mathbb{Z}.$$