I want to show that $\lambda \in \sigma (T)$ implies $|\lambda|\leq ||T||_{B(X)}$ where $\sigma (T)$ is the spectrum of $T$ and $T\in B(X)$. I would like to check my proof here as it is different and possibly simpler than the one in my notes.
Proof: Let $\lambda \in \sigma (T)$. This implies that $\ker(\lambda I-T)\neq 0$, so $\exists\ u \in \ker(\lambda I -T)$ where $u\neq 0$. Considering this $u$ we have $\lambda I(u) - T(u)=0 \implies \lambda u=T(u)$.Then, taking the norm of both sides, $||\lambda u||=||T(u)|| \implies |\lambda|||u||=||T(u)|| \implies |\lambda|=\frac {||T(u)||} {||u||}=||T(\frac{u}{||u||})||\leq||T||_{B(X)}$
Does this work?
You know from the comments that there was an error in your attempt, namely, that $\lambda$ need not be an eigenvalue. This need not even be true if $T$ is a self-adjoint operator on a Hilbert space.
The fact that the spectral radius is bounded by $\|T\|$ follows from the Neumann series representation of the resolvent and it is true on general Banach spaces.
Let $\lambda \in \mathbb{C}$ with $|\lambda|>\|T\|.$ Then $$\sum_{n=0}^{\infty} \frac{1}{\lambda^{n+1}}T^n$$ converges uniformly and absolutely and $\lambda \in \rho(T)$. In fact we have $$R(\lambda,T)=\sum_{n=0}^{\infty} \frac{1}{\lambda^{n+1}}T^n.$$ This implies $$\{\lambda \in \mathbb{C}:|\lambda|> \|T\|\}\subseteq \rho(T).$$ In other words $$\sigma(T)\subseteq\{\lambda \in \mathbb{C}:|\lambda|\leq \|T\|\}.$$