Denote $Y=\operatorname{Im}(T_\lambda)$, then $Y$ is a proper subspace of $H$. By projection theorem $Y^\perp\neq \{0\}$, so you can choose $y\in Y^\perp\setminus\{0\}$ which is by definition of $Y$ is orthogonal to $Y$, i.e. to $\operatorname{Im}(T_\lambda)$
The continuous spectrum of an operator may not look like an interval at all. In fact, it can very well be the Cantor set!
Here is an example:
Let $\mu $ be any positive Borel measure on the Cantor set, here denoted by $C$, such that $\mu $ has no atoms
(that is, $\mu (\{x\})=0$, for all $x$ in $C$) and full support
(that is, $\mu (U)>0$, for all nonempty
open sets $U\subseteq C$).
See below for the construction of such a measure.
Consider the operator $T$ on $L^2(C,\mu )$ given by
$$
T(\xi )|_x = h(x)\xi (x), \quad \forall \xi \in L^2(C,\mu ), \quad \forall x\in C,
$$
where $h$ is the function defined on $C$ by $h(x)=x$.
Using this wikipedia page it follows
that the spectrum of $T$ coincides with its continuous spectrum, which in turn coincides with the range of $h$, a.k.a $C$.
Here is one way to construct a measure $\mu $ on the Cantor set with the required properties. First of all recall that
the Cantor set is homeomorphic to $\{0,1\}^{\mathbb N}$, also known as Bernouli's space. For a concrete homeomorphism
take
$
\varphi :\{0,1\}^{\mathbb N}\to C,
$
given by
$$
\varphi (x) = \sum_{n=1}^\infty x_n3^{-n},
$$
for every $x=(x_1,x_2,\ldots ) \in \{0,1\}^{\mathbb N}$.
Consider the uniform probability measure $\rho $ on $\{0,1\}$, given by $\rho (\{0\}) =\rho (\{1\}) =1/2$, and let
$$
\nu =\prod_{n=1}^\infty \rho
$$
be the corresponding product measure. Incidentally $\nu $ is known as the Bernouli measure.
It is well known (and easy to prove) that $\nu $ has no atoms and full support. Since $\varphi $ is a homeomorphism, it
follows that the push forward measure $\mu := \varphi _*(\nu )$ has the same properties.
Best Answer
Being orthogonal makes no difference in terms of what the spectrum is. You have
if $P=I$, then $\sigma(P)=\{1\}$;
if $P=0$, then $\sigma(P)=\{0\}$;
if $P$ is not $0$ nor $I$, then $\{0,1\}\subset\sigma(P)$ (because $P(I-P)=0$ shows that neither can be invertible) and for any $\lambda\in\mathbb C\setminus\{0,1\}$ one can check directly that $$ (P-\lambda I)^{-1}=\frac1{\lambda(1-\lambda)}\,P-\frac1\lambda\,I. $$
In summary, if $P^2=P$ then
$\sigma(P)=\{1\}$ if and only if $P=I$;
$\sigma(P)=\{0\}$ if and only if $P=0$;
$\sigma(P)=\{0,1\}$ if and only if $P\ne 0,I$.
In all cases the spectrum agrees with the point spectrum. For if $x\in P\mathcal{H}$, then $Px=x $ and so $1$ is an eigenvalue.