Let $(Y_{i})_{i\in I}$ be an open affine cover of $V$ and set $V^{i}:=\red^{-1}(Y_{i})$. Then $V^{i}$ is strictly $K$-affinoid by [Bos77, Theorem 3.1] and hence $V^{i}_{x}$ is a good germ. Note that $(V^{i}_{x})_{i\in I}$ is a cover of $X_{x}$. Hence $\tilde{X_{x}}$ is obtained by glueing the $\tilde{V^{i}_{x}}$ along the canonical maps $\widetilde{V^{i}_{x}\cap V^{j}_{x}}\rightarrow\tilde{V^{i}_{x}}$. Let $V^{i}=\mathscr{M}(A_{i})$ for a strictly $K$-affinoid algebra $A_{i}$. Then $Y_{i}=\Spec{\tilde{A_{i}}}$ and the character $\chi_{x}:A_{i}\rightarrow\mathscr{H}(x)$ induces a morphism $\tilde{\chi_{x}}:\tilde{A_{i}}\rightarrow\widetilde{\mathscr{H}(x)}$. Let $\mathfrak{p}\subseteq\tilde{A_{i}}$ be the prime ideal corresponding to $\red(x)$ i.e. $\mathfrak{p}$ is the kernel of $\tilde{\chi_{x}}$. The induced morphism $\tilde{A_{i}}/\mathfrak{p}\rightarrow\widetilde{\mathscr{H}(x)}$ is injective and hence it extends to a morphism $\tilde{K}(V)\rightarrow\widetilde{\mathscr{H}(x)}$ where $\tilde{K}(V)=\Quot(\tilde{A_{i}}/\mathfrak{p})$ denotes the function field of $V$. This induces a morphism $\pi:\boldsymbol{P}_{\widetilde{\mathscr{H}(x)}/\tilde{K}}\rightarrow\boldsymbol{P}_{\tilde{K}(V)/\tilde{K}}$.
First step: We have that $\tilde{V^{i}_{x}}=\boldsymbol{P}_{\widetilde{\mathscr{H}(x)}/\tilde{K}}\{\tilde{\chi_{x}}(\tilde{A_{i}})\}$ is the preimage under $\pi$ of the set of valuation rings in $\tilde{K}(V)$ which admit a center on $Y_{i}\cap V$.
Indeed if $R\in\boldsymbol{P}_{\widetilde{\mathscr{H}(x)}/\tilde{K}}$ is a valuation ring with $\tilde{\chi_{x}}(\tilde{A_{i}})\subseteq R$ then $\tilde{A_{i}}/\mathfrak{p}\subseteq R\cap\tilde{K}(V)$. Let $\mathfrak{m}_{R}$ be the maximal ideal of $R$ then $\mathfrak{p}^{\prime}:=\mathfrak{m}_{R}\cap\tilde{A_{i}}/\mathfrak{p}$ defines a point in $\Spec(\tilde{A_{i}}/\mathfrak{p})$ whose local ring is $(\tilde{A_{i}}/\mathfrak{p})_{\mathfrak{p}^{\prime}}$ and we have $(\tilde{A_{i}}/\mathfrak{p})_{\mathfrak{p}^{\prime}}\subseteq R$. Then $R\cap\tilde{K}(V)$ admits the center $\mathfrak{p}^{\prime}$ on $Y_{i}\cap V$ as claimed. Conversely if $R\cap\tilde{K}(V)$ admits a center on $Y_{i}\cap V$ then there exists $\mathfrak{p}^{\prime}\in\Spec(\tilde{A_{i}}/\mathfrak{p})$ such that $(\tilde{A_{i}}/\mathfrak{p})_{\mathfrak{p}^{\prime}}\subseteq R\cap\tilde{K}(V)$ and hence obviously $\tilde{\chi_{x}}(\tilde{A_{i}})\subseteq R$.
Second step: The map $\tilde{X_{x}}\rightarrow\boldsymbol{P}_{\widetilde{\mathscr{H}(x)}/\tilde{K}}$ is surjective if and only if any valuation on $\tilde{K}(V)/\tilde{K}$ admits at least one center on $V$.
Let $\tilde{X_{x}}\rightarrow\boldsymbol{P}_{\widetilde{\mathscr{H}(x)}/\tilde{K}}$ be surjective and $v$ a valuation on $\tilde{K}(V)/\tilde{K}$. Then $v$ extends to a valuation $\tilde{v}$ on $\widetilde{\mathscr{H}(x)}$. Let $R$ be the valuation ring of $\tilde{v}$. Then $R\in\boldsymbol{P}_{\widetilde{\mathscr{H}(x)}/\tilde{K}}$ and hence $R$ has a preimage $R^{\prime}\in\tilde{X_{x}}$. Then there exists $i\in I$ such that $R^{\prime}\in\tilde{V^{i}_{x}}$ hence the image of $R^{\prime}$ in $\boldsymbol{P}_{\tilde{K}(V)/\tilde{K}}$ admits a center on $Y_{i}\cap V$ by the first step. But this image is $R\cap\tilde{K}(V)$ by construction which is the valuation ring of $v$. Hence $v$ admits a center on $V$. Conversely suppose that any valuation on $\tilde{K}(V)/\tilde{K}$ admits a center on $V$ and let $R\in\boldsymbol{P}_{\widetilde{\mathscr{H}(x)}/\tilde{K}}$ then the image of $R$ in $\boldsymbol{P}_{\tilde{K}(V)/\tilde{K}}$ induces a valuation on $\tilde{K}(V)/\tilde{K}$ which admits a center $z$ on $V$. Let $i\in I$ such that $z\in Y_{i}$ then $R\in\tilde{V^{i}_{x}}$ and the induced element in $\tilde{X_{x}}$ is a preimage of $R$.
Third step: The map $\tilde{X_{x}}\rightarrow\boldsymbol{P}_{\widetilde{\mathscr{H}(x)}/\tilde{K}}$ is injective if and only if every valuation on $\tilde{K}(V)/\tilde{K}$ admits at most one center on $V$.
To see this we describe $\widetilde{V^{i}_{x}\cap V^{j}_{x}}$. In order to do so we cover $Y_{i}\cap Y_{j}$ by open affine subsets $Y_{i,j}^{k}$. Their preimages under $\red$ yield a cover of $V^{i}_{x}\cap V^{j}_{x}$ by good germs. As above their reductions can be described as the preimage of the set of valuation rings in $\tilde{K}(V)$ which admit a center on $Y_{i,j}^{k}\cap V$. The reduction of $V^{i}_{x}\cap V^{j}_{x}$ is then obtained by glueing these spaces. Now suppose that any valuation on $\tilde{K}(V)/\tilde{K}$ admits at most one center and let $R_{1},R_{2}\in\tilde{X_{x}}$ which map to the same valuation ring $R\in\boldsymbol{P}_{\widetilde{\mathscr{H}(x)}/\tilde{K}}$. There exists $i,j$ such that $R_{1}\in\tilde{V^{i}_{x}},R_{2}\in\tilde{V^{j}_{x}}$. As we have seen in the first step, $R_{1}\cap\tilde{K}(V)$ and $R_{2}\cap\tilde{K}(V)$ admit centers $y_{1}\in\Spec(\tilde{A_{i}})\cap V$ respectively $y_{2}\in\Spec(\tilde{A_{j}})\cap V$. Then both are a center of $R\cap\tilde{K}(V)$. Hence $y_{1}=y_{2}\in Y_{i}\cap Y_{j}$ by our assumption. Therefore by the first step $R_{1}=R_{2}=R$ in $\widetilde{V^{i}_{x}\cap V^{j}_{x}}$. Hence in the glueing process, $R_{1}$ and $R_{2}$ are identified with each other. Conversely suppose that there is a valuation on $\tilde{K}(V)/\tilde{K}$ which admits two centers $y_{1},y_{2}\in V$. Let $y_{1}\in Y_{i}$ and $y_{2}\in Y_{j}$. Choose an extension of the valuation to $\widetilde{\mathscr{H}(x)}$ and let $R$ denote its valuation ring. Then $R$ induces an element $R_{1}\in\tilde{V^{i}_{x}}$ as well as an element $R_{2}\in\tilde{V^{j}_{x}}$. Then $R_{1}$ and $R_{2}$ map to the same element $R$ in $\boldsymbol{P}_{\widetilde{\mathscr{H}(x)}/\tilde{K}}$ but they are not identified in the glueing process as $Y_{i}\cap V$ and $Y_{j}\cap V$ are separated and hence $R_{1}$ and $R_{2}$ admit at most one center in $\Spec(\tilde{A_{i}})\cap V$ respectively $\Spec(\tilde{A_{j}})\cap V$ which means in particular that they do not admit a center in $Y_{i}\cap Y_{j}\cap V$. Hence $\tilde{X_{x}}\rightarrow\boldsymbol{P}_{\widetilde{\mathscr{H}(x)}/\tilde{K}}$ is not injective. This proves the third step.
Recall that $V$ is proper if and only if every valuation on $\tilde{K}(V)/\tilde{K}$ admits a unique center on $V$ ([Har77, Ch. II, Ex. 4.5]). Hence the claim follows from the second and third step. ∎