Let T: $\mathbb C^n \rightarrow \mathbb C^m$ be a linear map and W be a d-dimensional subspace of $\mathbb C^n$ such that the nullity of $T|_{w}$ is s. Then the nullity of T is at most n-d +s. Is this true or false?
Observations So the question leads me to try "decomposing" T into W and $W^c$, the orthogonal complement of W in T. So we know that W has s elements in the basis for the kernel of T. And we also know that $W^c$ has dimension of n-d. So the kernel of T in W^c has at most dimension n-d. Is it okay for me to then say that therefore the statement is true? Or did I do something wrong here?
The statement, and your explanation of the proof, are both correct but not completely clear.
It is not $T$ which is being decomposed, but rather $\mathbb C^n$ into $W$ and $W^c$.
If you want to be a little more formal, this is the way to do it : apply the rank nullity theorem to $T|_W$ to get $\mbox{ran } T|_W + \ker T|_W = d \implies \mbox {ran } T|_W = d-s$.
Now, $\mbox{ran } T|_W \leq \mbox{ran } T$. Thus, $\mbox{ran } T \geq d-s$, so $\ker T \leq n-d+s$, since the sum of these two must be $n$ by the rank-nullity theorem.
Note that the inequality cannot be made stronger. Can you see why?