Adjoints of closed densely-defined linear operators on a Hilbert space $X$ are nice, once you get used to working in the graph space. In fact, the proofs are easier using these techniques for general closed densely-defined operators than the special-case proofs offered for the bounded case. John von Neumann introduced this way of working with densely-defined linear operators on a Hilbert space $X$. I'll explain his approach.
A Graph: The first thing to observe is that a subspace $\mathcal{M}\subseteq X\times X$ is the graph $\mathcal{G}(L)=\{ \langle x, Lx\rangle : x\in\mathcal{D}(L)\}$ of a linear operator $L : \mathcal{D}(L)\subseteq X\rightarrow X$ iff $\langle 0,y\rangle \in \mathcal{M}$ implies $y=0$. And $L$ is a closed linear operator iff its graph is a closed in the product space $X\times X$. If the subspace $\mathcal{M}$ is a graph, then the domain of $L$ becomes the set of all first coordinates in the subspace $\mathcal{M}$. It is easy to show that this unique correspondence between $x \in \mathcal{D}(L)$ and the second coordinate is linear because $\mathcal{M}$ is linear.
Closable: A linear operator $L : \mathcal{D}(L)\subseteq X\rightarrow X$ is closable iff the closure $\mathcal{G}(L)^{c}$ in $X\times X$ of its graph $\mathcal{G}(L)$ is a graph. Equivalently, if $\{ x_{n} \}\subseteq \mathcal{D}(L)$ converges to $0$ and $\{ Ax_{n} \}$ converges to some $y$, then $y=0$. That's the condition for a linear operator to have a closed extension, and is equivalent to the requirement that the closure of the graph of $L$ be the graph of a linear operator.
Inverses: If $L$ is a linear operator, then $L^{-1}$ exists iff the transpose of the graph of $L$ is a graph. Using the transpose map $\tau\langle x,y\rangle = \langle y,x\rangle$,
$$
\tau \mathcal{G}(L)=\mathcal{G}(L^{-1}),
$$
provided both exist. Notice that if $L^{-1}$ exists, then $L^{-1}$ is closed iff $L$ is closed because $\tau$ is a unitary map on $X\times X$ with $\tau^{2}=I$.
Adjoints: The real power of using graphs comes in defining the adjoint. If $L$ is a closed densely-defined linear operator, then $L$ has a closed densely-defined adjoint $L^{\star}$ whose graph is
$$
\mathcal{G}(L^{\star})=J[\mathcal{G}(L)^{\perp}],
$$
where the orthogonal complement is taken in $X\times X$ and $J$ is the unitary symplectic transpose
$$
J\langle x, y\rangle = \langle y, -x\rangle.
$$
Notice that $J^{2}=-I$. Because $\tau$ and $J$ are unitary, they commute with the action of taking orthogonal complement. So you may also write
$$
\mathcal{G}(L^{\star})=J[\mathcal{G}(L)^{\perp}]=[J\mathcal{G}(L)]^{\perp}.
$$
Of course the transpose operator $\tau$ also commutes with the action of taking the orthogonal complement.
Commutativity: Also note that $\tau J = - J\tau$, which means $\tau J\mathcal{M}=J\tau\mathcal{M}$ for subspaces $\mathcal{M}$ of $X\times X$. So, when you are considering the action of $J$, $\tau$ and $\perp$ on subspaces $\mathcal{M}\subseteq X\times X$, you can freely interchange these operations. That makes life very simple.
Your Example: Suppose that $L$ is a closed densely defined linear operator with a densely-defined inverse $L^{-1}$. Automatically $L^{-1}$ is closed because $L$ is closed (their graphs are transposes of each other.) That means that $L^{-1}$ will have a closed densely-defined adjoint $(L^{-1})^{\star}$. As you might guess, $(L^{-1})^{\star}=(L^{\star})^{-1}$.
To show $(L^{-1})^{\star}=(L^{\star})^{-1}$: First, you must show that $L^{\star}$ has an inverse, which comes down to showing $\tau\mathcal{G}(L^{\star})$ is a graph:
$$
\tau\mathcal{G}(L^{\star})=\tau[J\mathcal{G}(L)^{\perp}]=
[J\tau\mathcal{G}(L)]^{\perp}=
[J\mathcal{G}(L^{-1})]^{\perp}=\mathcal{G}((L^{-1})^{\star}).
$$
Obviously the subspace on the far right is a graph. So $L^{\star}$ has an inverse and $(L^{\star})^{-1}=(L^{-1})^{\star}$. This proves the following:
Lemma: Let $H$ be a Hilbert space and $L$ a densely-defined closed linear operator on $H$. If $L$ has a densely-defined inverse $L^{-1}$, then $L^{\star}$ has a densely-defined inverse, and $(L^{\star})^{-1}=(L^{-1})^{\star}$.
Note: If $L^{-1}$ is defined everywhere then it is bounded by the closed graph theorem. In that case $(L^{\star})^{-1}$ is also defined everywhere and is bounded because of the graph equation stated in the lemma. This is the case in your problem for $L=U-\lambda I$ because resolvents are, by definition, defined everywhere and are bounded.
Added in Response to your Addition: Another big fact. If $A$ is closed and densely-defined then $A^{\star\star}=A$. This is because $(\mathcal{M}^{\perp})^{\perp}=\mathcal{M}$ for a closed subspace of a Hilbert space such as $H\times H$. This extends the previous lemma.
Lemma: Let $L$ be a closed densely-defined linear operator on a Hilbert space $H$ with adjoint $L^{\star}$. Then $L^{-1}$ exists as a densely-defined linear operator iff $(L^{\star})^{-1}$ exists as a densely-defined linear operator and, in either case, $(L^{-1})^{\star}=(L^{\star})^{-1}$.
Proof: I showed you that if $L$ has a densely-defined inverse, then $L^{\star}$ has a densely defined inverse and $(L^{-1})^{\star}=(L^{\star})^{-1}$. Conversely, if $L^{\star}$ has a densely-defined inverse, then $(L^{\star})^{\star}=L$ has a densely-defined inverse and $((L^{\star})^{-1})^{\star}=L^{-1} \implies (L^{\star})^{-1}=(L^{-1})^{\star}$. $\;\;\Box$
As a corollary: If $L$ is a closed densely-defined linear operator on a Hilbert space, then $L-\lambda I$ has a densely-defined inverse iff $L^{\star}-\overline{\lambda}I$ has a densely-defined inverse and, in either case,
$$
((L-\lambda I)^{-1})^{\star}=(L^{\star}-\overline{\lambda}I)^{-1}.
$$
In particular, $\lambda \in\rho(L)$ iff $\overline{\lambda}\in\rho(L^{\star})$ and the resolvents satisfy $R_{L}(\lambda)^{\star}=R_{L^{\star}}(\overline{\lambda})$. This last equation holds very generally in the sense that one exists iff the other does and, in that case, the two always equal. So, if $L=L^{\star}$ then $\lambda\in\rho(L)$ iff $\overline{\lambda}\in\rho(L)$ and, in that case, $R_{L}(\lambda)^{\star}=R_{L}(\overline{\lambda})$.
Best Answer
Note that the diagonal $\Delta=\{(x,x)\mid x\in[0,1]\}$ is a measure zero set, so it is not a priori clear what the meaning should be of a symbol such as $k(x,x)$. But we do have the following
Proof: Since $K$ is self-adjoint and compact, the spectral theorem ensures the existence of an ONB $(e_i)_{i\in\mathbb{N}}$ such that, for any $\psi\in L^2([0,1],\mathbb{C})$, we have $$ K\psi=\sum_{i=1}^\infty \lambda_i \langle e_i,\psi\rangle e_i. $$ Furthermore, since $K$ is Hilbert-Schmidt, we have $\sum_{i=1}^\infty \lvert \lambda_i\rvert^2<\infty$. Fixing representatives $e_i:[0,1]\rightarrow \mathbb{C}$ for the basis $(e_i)_{i\in\mathbb{N}}$, it follows that the series $$ q(x,y):=\sum_{i=1}^\infty \lambda_i e_i(x) \bar{e}_i(y) $$ is convergent in $L^2([0,1]^2,\mathbb{C})$, and since we have \begin{align} \int \bar{\phi}(x) k(x,y) \psi(y) \, dxdy & = \langle \phi , K\psi\rangle= \sum_{i=1}^\infty \lambda_i \langle \phi, e_i\rangle \langle e_i,\psi\rangle \\ &= \sum_{i=1}^\infty \lambda_i \int \bar{\phi}(x) e_i(x) \bar{e}_i(y) \psi(y)\, dxdy = \int \bar{\phi}(x) q(x,y) \psi(y)\, dxdy \end{align} for all $\phi,\psi\in L^2([0,1],\mathbb{C})$, it follows that $q=k$ almost everywhere.
Now we make use of the assumption that $K$ is trace class: We have $\sum_{i=1}^\infty \lvert \lambda_i\rvert<\infty$. Thus, we can define $k(x):=q(x,x)=\sum_{i=1}^\infty \lambda_i \lvert e_i(x)\rvert^2$, and then we have $k\in L^1([0,1],\mathbb{C})$. Furthermore, we have $$ \int k(x)\,dx = \sum_{i=1}^\infty \lambda_i \int \lvert e_i(x)\rvert^2\, dx = \sum_{i=1}^\infty \lambda_i = \mathrm{Tr }\, K, $$ finishing the proof.
This solution does not make use of non-negativity of K. To make the connection to the answer by Christian Remling, note that $K$ has a non-negative square root $A=\sqrt{K}$, and we have $$ A\psi=\sum_{i=1}^\infty \sqrt{\lambda_i} \langle e_i,\psi\rangle e_i. $$ Since $K$ is trace class, it follows that $A$ is Hilbert-Schmidt, and as above we find a kernel for $A$, namely $$ a(x,y)=\sum_{i=1}^\infty \sqrt{\lambda_i}e_i(x)\bar{e}_i(y). $$ Then we have $$ q(x,y) = \sum_{i=1}^\infty \lambda_i e_i(x)\bar{e}_i(y) = \int a(x,t)a(t,y) \, dt , $$ and in particular $k(x)=\int a(x,t)a(t,x) \, dt$. Here, it is also possible to remove the assumption that $K$ is non-negative, by noting that the polar decomposition $K=U\lvert K \rvert=(U\lvert K \rvert^{1/2})\lvert K \rvert^{1/2}$ affords us with a factorization of $K$ into a product of Hilbert-Schmidt operators.
Questions of this sort have received some attention in the literature, see for instance the paper 'Traceable Integral Kernels on Countably Generated Measure Spaces' by Brislawn. There is also a related question on MO, When is an integral transform trace class.
Brislawn, Chris, Traceable integral kernels on countably generated measure spaces, Pac. J. Math. 150, No.2, 229-240 (1991). ZBL0724.47014.