I have used the spectral theorem for normal operators, to prove some basic results related to the spectra of operators. Please provide feedback for my proofs, and feel free to suggest any other ways (using the spectral theorem).
The Spectral Theorem. If $T$ is a normal operator on a separable space, then there exists a finite measure space $(X,\mu)$ and $h\in L^\infty(X,\mu)$ such that $T$ is unitarily equivalent to the operator $M_h$ on $L^2(X,\mu)$.
Throughout, assume that $T$ is a normal operator. By the Spectral Theorem as stated above, there exists a unitary $U$ such that $U^*TU = M_h$. We have $\sigma(T) = \sigma(UM_hU^*) = \sigma(M_h) = \operatorname{essran} h$ (the essential range of $h$). I have also shown that if $\operatorname{essran} h \subset A \subset\mathbb C$, then $h(x) \in A$ a.e. on $X$, and I will use this in the proofs below.
$1$. $T$ is Hermitian if and only if $\sigma(T) \subset \mathbb R$.
Suppose $T$ is Hermitian, i.e. $T=T^*$. So, $M_h = M_h^* = M_{\overline{h}}$. This implies $h = \overline{h}$ everywhere, so $h$ is real. As $\operatorname{essran} h \subset \overline{\operatorname{im} h} \subset \mathbb R$, we have $\sigma(T) \subset \mathbb R$.
Conversely, suppose $\sigma(T) \subset \mathbb R$, i.e. $\operatorname{essran} h \subset \mathbb R$. So, $h$ is real a.e., i.e. $h = \overline{h}$ a.e. Let $f\in L^2(X,\mu)$. As $h = \overline{h}$ a.e., we have $hf = \overline{h}f$ a.e.; but elements of $L^2(X,\mu)$ are really equivalence classes of functions agreeing with each other a.e, so $M_h f = M_{\overline{h}} f$. That is, $M_h = M_{\overline{h}} = M_h^*$. As $T = UM_hU^*$, we have $T=T^*$.
$2$. $T$ is unitary if and only if $\sigma(T) \subset S^1$, where $S^1 := \{z\in \mathbb C: |z| = 1\}$.
Suppose $T$ is unitary, i.e. $TT^* = T^*T = I$. We get $TT^* = UM_hU^*UM_{\overline{h}}U^* = UM_hM_{\overline{h}} U^* = UM_{h\overline{h}}U^* = I$. So, $M_{h\overline{h}} = I$. Take arbitrary $f\in L^2(X,\mu)$ and see that $f = h\overline{h} f$ which means $|h|^2 = 1$ giving $|h| = 1$. That is, $\operatorname{im} h \subset S^1$. Since closure preserves inclusions, and $S^1$ is a closed subset of $\mathbb C$, I argue that $\operatorname{im} f\subset S^1$. As $\operatorname{essran} h \subset \overline{\operatorname{im} h}$, we have $\operatorname{essran} h \subset S^1$. Therefore, $\sigma(T)\subset S^1$.
Conversely, suppose $\sigma(T) \subset S^1$, i.e. $\operatorname{essran} h \subset S^1$. So, $|h| = 1$ a.e. on $X$, i.e. $h\overline{h} = 1$ a.e. Using an argument similar to that in the previous proof, we conclude $M_{h\overline{h}} = M_h M_{\overline{h}} = I$. It follows that $TT^* = T^*T = I$, i.e. $T$ is unitary.
$3$. $T$ is positive if and only if its spectrum is on the non-negative real axis, i.e. $\sigma(T) \subset \mathbb R_{\ge 0}$.
Firstly, let us see that $T$ is positive if and only if $M_h$ is positive. We have $\langle Tx,x\rangle = \langle UM_hU^*x,x\rangle = \langle M_hU^*x, U^*x\rangle$, and the claim follows from the observation that $U^* = U^{-1}$ is invertible.
Now, suppose $T\ge 0$, i.e. $M_h \ge 0$. So, for every $f\in L^2(X,\mu)$, $\langle M_hf,f\rangle \ge 0$. Expanding the inner product, we have $\langle M_hf, f\rangle = \int_X hf\overline{f}\, d\mu = \int_X h|f|^2 \, d\mu \ge 0$. Putting $f = \chi_{\{h < 0\}}$, we get $\int_X h\chi_{\{h < 0\}}\, d\mu = \int_{\{h < 0\}} h\, d\mu \ge 0$; but $\int_X h\chi_{\{h < 0\}}\, d\mu = \int_{\{h < 0\}} h\, d\mu \le 0$, giving $\int_{\{h < 0\}} h\, d\mu = 0$. This forces $\mu(\{h < 0\}) = 0$, i.e. $h\ge 0$ a.e. Define $h_1:= h\chi_{h\ge 0}$. Clearly, $h = h_1$ a.e., so $\operatorname{ess ran} h = \operatorname{ess ran} h_1$. As $\operatorname{ran} h_1 \subset [0,\infty)$ (giving $\overline{\operatorname{ran} h_1} \subset [0,\infty)$), we have $\operatorname{ess ran} h_1 \subset [0,\infty)$. It follows that $\operatorname{ess ran} h \subset \mathbb R_{\ge 0}$, as required.
For the converse, assume $\sigma(T)\subset \mathbb R_{\ge 0}$. Then, $\operatorname{essran} h \subset \mathbb R_{\ge 0}$. This implies $h(x) \ge 0$ a.e. on $X$. Therefore, for any $f\in L^2(X,\mu)$, $\int_X h|f|^2\, d\mu = \int_X hf\overline{f} d\mu \ge 0$, i.e. $\langle M_h f,f\rangle \ge 0$. So, $M_h \ge 0$ which yields $T\ge 0$.
$4$. $T$ is a projection if and only if $\sigma(T)\subset \{0,1\}$.
If $P\in \mathcal B(\mathcal H)$ is an idempotent operator, i.e. $P^2 = P$, then the following are equivalent: (i) $P$ is an orthogonal projection onto a closed subspace (ii) $P$ is Hermitian (iii) $P$ is normal.
Suppose $T$ is a projection; then $T^2 = T$. So, $M_{h^2} = M_h$ which gives $h = h^2$ a.e., i.e. $h(x) \in \{0,1\}$ a.e. Define $h_1:= h\chi_{S}$, where $S = h^{-1}(\{0,1\})$. Clearly, $h_1 = h$ a.e. and $h_1(x) \ge 0$ for all $x\in X$. Consequently, $\operatorname{ess ran} h = \operatorname{ess ran} h_1 \subset \overline{\operatorname{ran} h_1} \subset \{0,1\}$. Thus, $\sigma(T) \subset \{0,1\}$.
Conversely, suppose $\sigma(T)\subset\{0,1\}$, i.e. $\operatorname{ess ran} h \subset \{0,1\}$. Then, $h(x) \in \{0,1\}$ a.e., i.e. $h^2 = h$ a.e. which means $M_h = M_{h^2}$, and $T=T^2$. Since $\sigma(T) \subset \mathbb R$, $T$ is also Hermitian. In view of the aforementioned equivalence, $T$ is a projection.
Thank you very much for reading! I would appreciate any feedback.