I understand the definition of self-adjoint in terms of the inner product, but I am not sure how to do this as we are not given the inner product for the operator.
I'm sure that I have to use the norm at some point, since $u$ is a unit vector, but again not sure how I can make use of this without knowing the inner product.
On
Possibly a good starting point is to recall the defining property of the adjoint $T^*$ of an operator $T$ $$\langle\,Tx\mid y\,\rangle\;=\;\langle\,x\mid T^*y\,\rangle\quad\forall x,y\in H.$$ It involves the given operator and the scalar product while $x$ and $y$ run through all of $H$.
Now apply this to a rank-one operator $T$, thus $Tx=\langle x|u\rangle\,v\,$ with fixed $\,u,v\in H$, to find its adjoint: $$\langle\,Tx\mid y\,\rangle\;=\; \langle\,\langle x| u\rangle v\mid y\,\rangle\;=\; \langle x| u\rangle\:\langle v|y\rangle\;=\; \langle\,x\mid \langle y|v\rangle u\,\rangle \quad\forall x,y\in H,$$ hence $T^*y=\langle y|v\rangle\,u$. Compared with $T$, the vectors $u$ and $v$ are swapped.
In particular $T$ is self-adjoint if $\,v=u$.
Note that if $u$ is a unit vector then $x\mapsto\langle x|u\rangle u\,$ is the orthogonal projector onto the subspace $\mathbb Cu$.
Direct computation gives \begin{align*} \left<(I-2T)x,y\right>=\left<x,y\right>-2\left<x,u\right>\left<u,y\right> \end{align*} and \begin{align*} \left<x,(I-2T)y\right>=\left<x,y\right>-2\left<u,y\right>\left<x,u\right>. \end{align*}