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Let $X$ be a quasi-compact scheme. There exists a natural map $X\rightarrow\mathrm{Spec}(\mathcal{O}_X(X))$. Prove that the image of this map is dense.

I think I've got it under assumption $X$ is integral. If so then it is equivalent to proving that the generic point is mapped to the generic point. Suppose that $X$ is covered by finitely many $\mathrm{Spec} A_i$ and that the generic point is inside $\mathrm{Spec} A_j$ for $j=1,\dots ,k$. Then it is the generic point of $\mathrm{Spec} A_j$ and we show it is mapped to the generic point of $\mathrm{Spec}(\mathcal{O}_X(X))$. For suppose that for every $1\leq j\leq k$ we have the kernel of the map is a non-zero prime ideal $\mathfrak{p}_j$. Observe that the intersection of $\mathfrak{p}_j$ cannot be zero since $\mathcal{O}_X(X)$ is an integral domain. Hence $\exists b\neq 0$ that is in every $\mathfrak{p} _j$. Then the image of $b$ in $A_j$ is its restriction on $\mathrm{Spec} A_j$ hence $b=0$, a contradiction.

We can also assume $X$ is reduced, since $X_\mathrm{red}\rightarrow X$ is homeomorphism and idem for $\mathrm{Spec} (\mathcal{O}_{X_\mathrm{red}})\rightarrow\mathrm{Spec}\mathcal{O}_X$.

How to tackle the remaining case, that is $X$ is finite union of integral schemes?

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    A related question you might consider first is if $\phi:A\to B$ is a ring map with $A$ reduced, then the corresponding $Spec(B)\to Spec(A)$ has dense image if and only if $\phi$ is injective – 2017-02-10
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    One more thought: you may want to attempt to do your original question by arguing every open on $Spec(O_X(X))$ intersects the image of $X$. Of course you can reduce to distinguised affine opens of $Spec(O_X(X))$ – 2017-02-10

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Theorem: If $f:\mathrm{Spec}(B)\to \mathrm{Spec}(A)$ morphism of affine schemes, and $\varphi:A\to B$ the corresponding morphism of global sections. Then $\overline{\mathrm{Im}f}=V(\ker\varphi)$

Using this theorem:

Let $X=\bigcup_{i=1}^nU_i$, where $U_i=\mathrm{Spec}(A_i)$ affine open. And denote $f:X\to \mathrm{Spec}(A)$ the natural morphism, where $A=\mathcal{O}_X(X)$.

Let $f_i: U_i\to \mathrm{Spec} A$ (the restriction of $f$ to $U_i$), the corresponding map on the rings is the restriction map: $\varphi:A\to A_i$.

No $\overline{\mathrm{Im}f}=\bigcup_{i=1}^n\overline{\mathrm{Im}f_i}$.

But $\overline{\mathrm{Im}f_i}=V(\ker\varphi_i)$ , it follow that $ \overline{\mathrm{Im}f}=\bigcup_{i=1}^nV(\ker\varphi_i)=V(\bigcap_{i=1}^n\ker\varphi_i)$ .

If $a\in \bigcap_{i=1}^n\ker\varphi_i\subset A$, then $a|_{U_i}=\varphi_i(a)=0$ (that is the restriction of the global section $a$ at each open $U_i$ is zero), then $a=0$.

Hence $\overline{\mathrm{Im}f}=V(0)=\mathrm{Spec}(A)$.