Proof that the rank of a skew-symmetric matrix is at least $2$

For a skew symmetric (real) matrix, the eigenvalues are all purely imaginary. This is because if $Av = \lambda v$, then we have $\lambda \langle v,v\rangle = \langle \lambda v, v\rangle = \langle Av,v\rangle = \langle v, -Av \rangle = \langle v, -\lambda v\rangle = -\overline{\lambda} \langle v,v\rangle$, so we conclude that $\lambda = -\overline{\lambda}$, i.e., that $\lambda$ is purely imaginary. Here, we're using an Hermitian inner product.

For a real matrix, complex eigenvalues come in conjugate pairs, so the rank must be even.

There exists an invertible matrix $P$ such that $^t P A P$ is diagonal with blocks equal to $\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}$ or $0$ (it is a simple exercise in bilinear forms), so that the rank of $A$ is necessarily even.

Edit: The following answers the first part of the OP's question, without using the concept of eigenvalues. It works on all fields (including $\mathbb{R}$) with characteristic $\ne2$.

Every rank-$1$ matrix can be written as $A=uv^\top$ for some nonzero vectors $u$ and $v$ (so that every row of $A$ is a scalar multiple of $v^\top$). If $A$ is skew-symmetric, we have $A=-A^\top=-vu^\top$. Hence every row of $A$ is also a scalar multiple of $u^\top$. It follows that $v=ku$ for some nonzero scalar $k$. But then $vu^\top=-uv^\top$ implies that $kuu^\top=-kuu^\top$ or $2kuu^\top=0$, which is impossible because both $k$ and $u$ are nonzero and the characteristic of the field is not $2$. Therefore, skew-symmetric matrices cannot be rank-1 matrices, and vice versa.

When the underlying field has characteristic 2, the notions of symmetric matrices and skew-symmetric matrices coincide. Hence every nonzero matrix of the form $uu^\top$ with nonzero vector $u$ is a rank-1 skew-symmetric matrix.