Morphisms from a proper scheme to the affine line over a field must be constant.

First we claim that the image of $\varphi$ cannot be all of $\Bbb{A}^1$. If it were all of $\Bbb{A}^1$, by composing with $\Bbb{A}^1 \hookrightarrow \Bbb{P}^1$ we get a dominant map $X \to \Bbb{P}^1$ that is furthermore proper and thus has closed image. But now this means the map $X \to \Bbb{P}^1$ is surjective, a contradiction.

So we know the image of $\varphi$ is a proper closed subset of $\Bbb{A}^1$. It is now enough to observe that the image of a connected set under a continuous function is also connected. For then the only connected proper closed subset of $\Bbb{A}^1_k$ is a one point set.

Added: Why is $X \to \Bbb{P}^1$ proper?

Well this comes from the famous "Property $\mathscr{P}$" exercise:

Let $\mathscr{P}$ be a property of morphisms that is stable under composition and base extension. Then suppose $f : X\to Z$ which factors as $X \stackrel{g}{\to} Y \stackrel{h}{\to} Z$. If $f$ and the diagonal $\Delta{h}$ have property $\mathscr{P}$, then so does $g$.