We will use the properties $\det A=\det(A^t)$ and $\det(-A)=(-1)^d\det A$, where $d$ is the dimension of the matrix.
We have $\det(A+I)=\det(A^t+I)=\det(I-A)$. Let $x$ be such that $(A-I)x=0$. Then $\langle x,x\rangle=\langle x,Ax\rangle=\langle A^tx,x\rangle=-\langle Ax,x\rangle=-\langle x,x\rangle,$ which proves that $x=0$. We have seen that if $(A-I)x=0$ then $x=0$, hence $A-I$ is invertible. This implies that $\det(I-A)\neq 0 $. Since $\det(I-A)=\det(A+I)$ we get that $\det(A+I)\neq 0$, which proves that $A+I$ is invertible.
Using the mentioned properties, for a skew-symmetrix matrix of odd dimension the determinant is $0$.
In dimension 2, a non-zero skew-symmetric matrix is of the form $A:=\pmatrix{0&x\\-x&0}$ where $x\neq 0$. Its determinant is $x^2\neq 0$ hence $A$ is necessarily invertible. For the other even dimensions (say $d=2N$), we can construct a non-zero skew-symmetric matrix which is not invertible. Indeed, consider the block matrix defined by $ \pmatrix{A&0_{2,2N-2}\\ 0_{2N-2,2}& 0_{2N-2,2N-2} } $ where $0_{i,j}$ denoted the matrix with $i$ rows and $j$ columns and all its entries are zero as $A= \pmatrix{0&1\\-1&0}$.