I can't quite figure this out. Perhaps it's simple. 


If $S$ is an affine scheme over $\mathbb{C}$ and $s$ and $t$ are $\mathbb{C}$ points in the same connected component, is there a finite type connected $\mathbb{C}$ scheme mapping to $S$ such that $s$ and $t$ are contained in the image of this map? 



Connected components of infinite type schemes
That is false, and the suggestions by Ariyan Javanpeykar and Joe Berner lead to a counterexample. Let $R\_0$ be $\mathbb{C}[t]$. Let $\Sigma$ be the multiplicative closed set of all elements $u(t)\in \mathbb{C}[t]$ such that $u(0)u(1)\neq 0$. Let $R$ be the fraction ring $\Sigma^{-1}R\_0$. This is $\mathbb{C}$-algebra that is not of finite type. 


For $\lambda=0,1$, denote by $e\_\lambda$ the $\mathbb{C}$-algebra homomorphism, $$e\_\lambda:R\_0 \to \mathbb{C}, \ \ e\_\lambda(u) = u(\lambda).$$ This maps every element in $\Sigma$ to an invertible element of $\mathbb{C}$. Thus, there is a unique induced $\mathbb{C}$-algebra homomorphism, $$ f\_\lambda:R\to \mathbb{C},$$ with $f\_\lambda(u)$ equal to $e\_\lambda(u)$ for every $u\in R\_0$. The kernel of $f\_\lambda$ is a maximal ideal $\mathfrak{m}\_\lambda$, i.e., it is a $\mathbb{C}$-point of $\text{Spec}\ R$. Of course $\text{Spec}\ R$ is connected, and even integral. The only prime ideals are $\mathfrak{m}\_0$, $\mathfrak{m}\_1$ and the zero ideal. Indeed, the primes of $R$ are in bijective correspondence with the primes of $R\_0$ that are disjoint from $\Sigma$. The complement of $\Sigma$ equals the union of the prime ideals $\langle t\rangle$ and $\langle t-1\rangle$. By Prime Avoidance, every prime contained in this union is contained in one of these primes. The only prime ideals of $R\_0$ contained in one of these primes are the primes themselves and $\{0\}$. 


For every finite type $\mathbb{C}$-algebra $A$ that is an integral domain, for the group of units $A^\times$, the quotient group $A^\times/\mathbb{C}^\times$ is a finitely generated, Abelian group. However, the group $R^\times/\mathbb{C}^\times$ is infinitely generated. Thus, for every $\mathbb{C}$-algebra homomorphism, $$g:R\to A,$$ the induced map of groups of units must have a kernel, i.e., there exist nonscalar units $v\in R$ such that $g(v)$ equals $g(c)=c$ for some $c\in \mathbb{C}^\times$. Thus, the kernel of $g$ contains the nonzero element $v-c$. Since the kernel of $g$ is a prime ideal of $R$, the kernel equals $\mathfrak{m}\_0$ or $\mathfrak{m}\_1$. So the homomorphism $g$ factors through $f\_0$ or $f\_1$. Thus the $\mathbb{C}$-morphism $$\text{Spec}(g):\text{Spec}\ A \to \text{Spec}\ R,$$ is constant with image $\mathfrak{m}\_0$ or $\mathfrak{m}\_1$.


More generally, every finite type $\mathbb{C}$-algebra $A$ for which $\text{Spec}\ A$ is connected is a finite union of irreducible components. For every $\mathbb{C}$-morphism, $$h:\text{Spec}\ A \to \text{Spec}\ R,$$ the restriction of $h$ to every irreducible component is constant with image $\mathfrak{m}\_0$ or $\mathfrak{m}\_1$. Thus, the $h$-inverse images of the two open subsets, $$U\_\lambda=\text{Spec}\ R \setminus\{ \mathfrak{m}\_\lambda \}, \ \ \lambda=0,1,$$ give an open covering of $\text{Spec}\ A$. Since $\text{Spec}\ A$ is connected, one of these two open sets equals all of $\text{Spec}\ A$, i.e., also $h$ is constant. 


