 Solution

A Euclidean lattice in ${\mathbb{R}}^2$ is a discrete additive subgroup, equivalently the integer span of zero, one or two real-linearly independent vectors. Its rank is the number of these generators, or the dimension of its real span. Discreteness makes its intersection with any bounded set finite, so shortest nonzero vectors exist.

First $\Lambda \cap \mathbb{R}{w}_1=\mathbb{Z}{w}_1$. Otherwise an element $t{w}_1$ with noninteger $t$ could be reduced by its nearest integer multiple of $w_1$ to a nonzero vector of length at most $|w_1|/2$, contradicting minimality. Therefore $w_2\notin \mathbb{Z}{w}_1$ is not parallel to $w_1$.

Write any $v\in \Lambda$ as $v=a{w}_1+b{w}_2$ with real coefficients. Subtract nearest integer multiples to obtain $r=s{w}_1+t{w}_2\in \Lambda$, with $|s|,|t|\le 1/2$. If $t\ne 0$, it is not on the first axis, yetwith strict inequality because the independent vectors cannot achieve equality in the triangle inequality; if one coefficient vanishes the inequality is also strict. This contradicts minimality of $w_2$. If $t=0$ and $r\ne 0$, then $|r|\le |w_1|/2$, contradicting minimality of $w_1$. Thus $r=0$ and both original coefficients are integers. Hence $\boxed{\Lambda =\mathbb{Z}{w}_1+\mathbb{Z}{w}_2}$.

For $w_1=(1,0)$ and $w_2=(x,y)$, minimality implies $x^2+y^2\ge 1$ and $|w_2|\le |w_2\pm w_1|$, hence $|x|\le 1/2$. The exact region isThese conditions are sufficient: in the generated lattice, vectors with second coefficient $\pm 1$ have smallest length when the first coefficient is zero, allowing ties at $|x|=1/2$. A vector with second coefficient of magnitude at least two has length at least $2|y|>\sqrt{x^2+y^2}$, since $|y{|}^2\ge 3/4$ and $x^2\le 1/4$. Vectors on the first axis have length at least one. Thus the shaded region includes exactly all allowable choices, including ties on its boundary.

For a second lattice basis, write $v_1=a{w}_1+b{w}_2$, $v_2=c{w}_1+d{w}_2$ with integer coefficients, as just proved. Conversely each $w_i$ is an integer combination of the $v_j$. The change-of-basis matrix $\left(\begin{array}{cc}a& c\\ b& d\end{array}\right)$ and its inverse therefore both have integer entries. Their determinants are integers with product one, giving $\boxed{ad-bc=\pm 1}$. Thus lattice bases are related by a unimodular matrix.