Unique expression of a tensor as sum of two other tensors. Given a covariant tensor $\alpha$, how do I prove that it has a unique expression of form $\alpha = \beta + \gamma$ where $\beta$ is alternating and $A\gamma = 0, A$ being the alternation of $\gamma$? 
Currently I'm considering $\alpha = \beta$ when $\alpha$ is alternating and $\beta = A\alpha$ in general with $\gamma =$ Sym$\alpha$. Am I on the right track?
Another approach I'm considering is: take $\beta$ to be the linear combination of the basis elements of the alternating subspace and $\gamma$ to be the linear combinations of the elements in the basis of $T^r(V)$ but not in the basis of the alternating subspace. 
 A: If $A$ is the alternating map, then you can prove it is a projection, because $A^2=A$. For you to have this, it is necessary to have the $1/k!$ factor in the alternation.
Let's work on a single vector space, $V$, which is real and $n$ dimensional. Let $T^*(V)$ be the covariant tensor algebra of $V$ (the same is true for contravariant tensors too, though), which is a graded algebra $$ T^*(V)=\oplus_{i=0}^\infty T^{0,i}(V), $$ where $T^{0,k}(V)$ is the space of $k$th order covariant tensors.
If $A$ is defined as $$ A=\frac{1}{k!}\sum_{\sigma}\text{sgn}(\sigma)P_\sigma, $$ where $k$ is the grade of the subspace of $T^*(V)$ $A$ acts upon, then $A$ will be a projection.
It is a well known result from linear algebra, that a projection operator splits a vector space into the direct sum $$ V=\text{Im}(P)\oplus\ker(P), $$ where $P$ is a projection, with $P$ being the projection onto $\text{Im}(P)$ and $1-P$ being the projection onto $\ker(P)$.
This is unique, because the direct sum decomposition is unique and because of $P^2=P$.
Now if $\omega\in T^{0,k}(V)$, a $k$th order covariant tensor, then we have $$ \omega=A\omega+(1-A)\omega, $$ which satisfies your requirements and is unique by the uniqueness of the direct sum decomposition.
(Of course, the direct sum decomposition is $$ T^{0,k}(V)=\Lambda^k(V^*)\oplus N^k, $$ where $N^k=\ker(A)$.)
Note: You approach with $\text{Sym}(\omega)$ would not work, because it is true only for second order tensors that $T^{0,2}=\Lambda^2(V^*)\oplus S^2(V^*)$. The symmetry properties of higher order tensors are more complicated, because only for second order tensors do we only have total symmetry or antisymmetry. For a higher order tensor, it is possible that it is symmetric only under the exchange of say two arguments. Then said tensor is not symmetric, but the alternating map will still kill it. The subspace $N^k$ in my approach is precisely the space of all $k$th order covariant tensors that have at least one symmetric pair of arguments.
A: 
Currently I'm considering $\alpha = \beta$ when $\alpha$ is alternating and $\beta = A\alpha$ in general with $\gamma =$ Sym$\alpha$. Am I on the right track?

Nearly. Instead of taking $\gamma =\mathrm{Sym}\alpha$ just take $\gamma=\alpha-\beta$ or in other words $\gamma=\alpha-A\alpha$. Then we automatically have $\alpha=\beta+\gamma$, and we have $A\gamma=A(\alpha-A\alpha)=A\alpha-AA\alpha=A\alpha-A\alpha=0$, which is what we wanted.
