I want to show that $X:=\text{Spec}(\mathbb{C}\times \mathbb{C})$ is a projective scheme. So we have to find a closed immersion from $X$ to $\mathbb{P}^{1}_{\mathbb{C}}$.
Brief explanation of my approach: I first tried to find a map $i:X\rightarrow\mathbb{P}^{1}_{\mathbb{C}}$ by noticing that $X$ can be covered by the opens $V_{1} = \{[\mathbb{C}\times\{0\}]\}$ and $V_{2} = \{[\{0\}\times\mathbb{C}]\}$. Now consider the standard open affine cover of $\mathbb{P}^{1}_{\mathbb{C}}$ given by $U_{1}=\text{Spec}(R_{0})$, where $R_{0} = \mathbb{C}[X_{10}]$ and $U_{2}=\text{Spec}(R_{1})$, where $R_{1} = \mathbb{C}[X_{01}]$.
Now I want to find morphisms $f_{i}: V_{i}\rightarrow U_{i}$ and glue them together to a morphism $f:X\rightarrow \mathbb{P}^{1}_{\mathbb{C}}$. Notice that to give a morphism $f_{i}$ is equivalent with giving a morphism $f_{i}^{*}:R_{i}\rightarrow \Gamma(V_{i},\mathcal{O}_{V_{i}})$.
First encountered problems: Now I struggle with determining what the global sections $\Gamma(V_{i},\mathcal{O}_{V_{i}})$ are, and how I can glue the morphisms in the end. My guess for $\Gamma(V_{i},\mathcal{O}_{V_{i}})$ would be that it is equal to $\mathbb{C}$, since then we would get a natural map for the $f_{i}^{*}$.
Here is an easier way to do this. Take a pair of distinct closed points $P,Q\in \mathbb{P}^n_{\mathbb{C}}.$ Then $X=\{P,Q\}\subseteq \mathbb{P}^n_{\mathbb{C}}$ is a closed, affine subscheme. Indeed, $\{P,Q\}$ admits an open cover by $\{P\}\cup \{Q\}$, and the functions on $\{P\}$ are constants and likewise for $\{Q\}$. Using the gluing axiom for sheaves, it follows that the global sections are isomorphic to $\mathbb{C}\times \mathbb{C}$, the identification being given by $f\in\mathcal{O}_X(X)\mapsto(f(P),f(Q)).$ Then, it is not hard to check that indeed $\operatorname{spec}(\mathbb{C}\times \mathbb{C})\cong \{P,Q\}$ so that $X$ is affine as claimed.
Then, the scheme $\operatorname{spec}(\mathbb{C}\times \mathbb{C})$ is projective, since it is isomorphic to a closed subscheme of $\mathbb{P}^n_{\mathbb{C}}$ obtained by choosing any pair of $P,Q$ distinct as above. If you wish, you can check that the map $\operatorname{spec}(\mathbb{C}\times \mathbb{C})\to\mathbb{P}^n_{\mathbb{C}}$ is a closed immersion, but it follows immediately from the above discussion.
An alternative approach: since you seem interested in gluing, we can do it that way too. We have an open cover of $\operatorname{spec}(\mathbb{C}\times \mathbb{C})$ by two copies of $\operatorname{spec}(\mathbb{C})$, which we can call $U_1$ and $U_2$. As you remarked in the comments above, $\operatorname{spec}(\mathbb{C}\times \mathbb{C})=U_1\sqcup U_2\cong \operatorname{spec}(\mathbb{C})\sqcup \operatorname{spec}(\mathbb{C}).$ To glue a pair of morphisms $f_1:U_1\to Y$ and $f_2:U_2\to Y$, it suffices to check that the morphisms $f_1$ and $f_2$ agree on the overlap $U_1\cap U_2$. Luckily, here $U_1\cap U_2=\varnothing$, so there is nothing to check. In particular, specifying $f:X\to \mathbb{P}^n$ is equivalent to specifying a pair of maps $$(f_1:U_1\to \mathbb{P}^n,f_2:U_2\to \mathbb{P}^n).$$ So, to do that we can choose a pair of closed points $P$ and $Q$ in $\mathbb{P}^n$. Then, let $f_1$ be the unique map sending $U_1$ to $\{P\}$ and $f_2$ the map sending $U_2$ to $\{Q\}$. This gives a morphism $X\to \{P,Q\}\subseteq \mathbb{P}^n$. You can now verify that this is a closed immersion.