Let $\mathbb{V}$ be a finite-dimensional inner product space, and let $\mathbb{W} \subset \mathbb{V}$ be a subspace.
Define $T:\mathbb{V} \rightarrow \mathbb{V}$ by $$T(\overrightarrow v)=\overrightarrow v + Proj_{W}\overrightarrow v$$
Show that $T$ is invertible.
My approach is to show that $T$ is injective and since $T$ goes from $\mathbb{V}$ to $\mathbb{V}$, that would imply that it is bjective and therefore invertible.
I let an arbitrary vector $\overrightarrow v \in KerT$.
$T(\overrightarrow v)=\overrightarrow 0$
$\Rightarrow$ $\overrightarrow v=-Proj_{W}\overrightarrow v$
This would mean that $\overrightarrow v$ is a linear combination of vectors which pertain to a basis of $\mathbb{W}$. $\therefore Proj_{W}\overrightarrow v=\overrightarrow v$
$\therefore \overrightarrow v=-\overrightarrow v \Rightarrow \overrightarrow v=\overrightarrow 0$
Therefore $T$ is bijective and invertible.
Is my approach correct?
Your work is correct (though only in finitely many dimensions, as is the case). It would be less awkward to say that $\overrightarrow v=-Proj_{W}\overrightarrow v$ implies that $\overrightarrow v \in \mathbb W$ than to say that it is "a linear combination of vectors which pertain to a basis of $\mathbb W$", but either is correct.
One may also note that a more direct proof would be to note that the inverse is $$T^{-1}(\overrightarrow v)=\overrightarrow v - \frac{1}2Proj_W\overrightarrow v$$ which can be shown to be the inverse directly by composing it on either side of $T$. This works in any number of dimensions, which is a plus.