Why are smooth/differentiable maps between manifolds continuous? I am currently doing one of the exercises in Michael Spivak's A Comprehensive Introduction to Differential Geometry: Volume 1, and it is written like so:

Prove that every $C^{\infty}$ function is continuous.

Let $f:M\to N$ for two $n$ and $m$-dimensional manifolds $M$ and $N$.  According to the text, Spivak defines $f$ to be $C^{\infty}$ (or differentiable) if for every coordinate system $(x,U)$ for $M$ and $(y,V)$ for $N$ that
$$y\circ f\circ x^{-1}:\mathbb{R}^n\to \mathbb{R}^m$$
is differentiable (which I assume to mean in the usual sense?).  Spivak says that if $f$ is $C^{\infty}$ then it is "clearly" continuous, and I had an initial argument which I hoped would show that this somehow was "clearly" true.
Proof: Let $f$ be $C^{\infty}$.  Then  for every such $(x,U)$ and $(y,V)$, the composition $y\circ f\circ x^{-1}:\mathbb{R}^n\to \mathbb{R}^m$ is $C^{\infty}$ in the usual sense, which means that $y\circ f\circ x^{-1}$ is continuous.  Since $y\circ f\circ x^{-1}$ is continuous (and $x$ and $y$ are assumed to be homeomorphisms) then this forces $f$ to be continuous, since the composition of continuous functions is continuous $\square$
I am a little skeptical of the above proof, mostly because it feels too good to be true.  However, I am not sure what else to try or how Spivak must have thought this was immediate.  Why is every $C^{\infty}$ function continuous? Thank you for your help, and my apologies if this is a stupid question.
 A: Alas, I have forgotten to fix my above proof with the assistance in the comments.  To fix the above proof, I will use the idea of @Mariano Suárez-Álvarez and look at this locally at some point in the domain $M$.
Suppose $M$ and $N$ are $C^{\infty}$ manifolds (with or without boundary), and let $f:M\to N$ be $C^{\infty}$.  Then for every $p\in M$, by the definition of smoothness there are smooth charts $(x, U)$ which contain $p$ and $(y, V)$ which contain $f(p)$ such that $f(U)\subseteq V$ and the local representation
$$\varphi= y\circ f\circ x^{-1}:x(U)\to y(V)$$
is $C^{\infty}$ in the normal sense.  This means that $\varphi$ is continuous (because differentiability implies continuity in the normal sense), and because $x:U\to x(U)$ and $y:V\to y(V)$ are assumed to be homeomorphisms, the restriction
$$f\big |_U=y^{-1}\circ (y\circ f\circ x^{-1})\circ x:U\to V$$
is the composition of continuous maps, which makes $f\big |_U$ continuous as well.  Because $f$ is continuous in a neighborhood $U$ of every point $p\in M$, it follows that $f$ is continuous and the proof is complete.
