Here's a proof using only basic trigonometry and algebra, no calculus or infinite series required.

We'll do a proof by contradiction. Suppose $\cos(x)$ is a polynomial on some closed interval $[a,b]$, with $a\ne b$. We'll split it into two cases, depending on whether or not $0\in [a,b]$.

**Case 1.** Suppose your interval contains the origin, i.e. $a \le 0 \le b$. If $\cos(x)$ is a polynomial function on $[a,b]$, then $2\cos^2(\frac x 2) - 1$ is also a polynomial function on $[a,b]$, since $x\in[a,b]$ implies $x/2 \in [a,b]$. Now, recall the half angle formula for $\cos(x)$:$$
\cos(x) = 2\cos^2(\frac x 2) - 1
$$ The half-angle formula tells us that these two polynomials are in fact *the same* polynomial. But if $\cos(x)$ has degree $n$, then $2\cos^2(\frac x 2) - 1$ must have degree $2n$. Since two polynomials with different degree cannot be equal on any interval, this implies $2n = n$, or $n=0$. Since $\cos(x)$ is not constant, we have a contradiction, so $\cos(x)$ is not a polynomial on any interval containing $0$.

**Case 2.** Now, what if the interval does not contain the origin? This takes a few more steps, but we can show that if $\cos(x)$ is a polynomial on $[a,b]$, then it must also be a polynomial (potentially a different polynomial) on $[0,b-a]$, which contains the origin so is impossible by the above argument.

For $x\in [0,b-a]$, we use the angle sum formula to find $$
\cos(x) = \cos(x+a -a) = \cos(x+a)\cos(a) + \sin(x+a)\sin(a)
$$
Since $\cos(x+a)$ is a polynomial of $x$, and $\sin(x+a)^2 + \cos(x+a)^2= 1$, this means that on the interval $[0,b-a]$, the cosine of $x$ has the property that $$
\left(\cos(x) - p(x)\right)^2 = q(x)
$$
for some polynomials $p$ and $q$. In particular $p(x) = \cos(a+x)\cos(a)$ and $q(x) = \sin^2(a) \left(1-\cos^2(x+a)\right)$. Equivalently, $\cos(x) = p(x) \pm \sqrt{q(x)}$.
Again, the half-angle formula tells us $\cos x = 2\cos^2(\frac x 2) - 1$ (for $x\in[0,b-a]$). Substituting into the above, we get some very messy algebra:\begin{eqnarray}
\left(2\cos^2\left(\frac x 2\right) - 1 - p(x)\right)^2 &=& q(x)\\
\left(2p(\frac x 2)^2 \pm 4 p(\frac x 2)\sqrt{q(\frac x 2)} + 2q(\frac x 2) - 1 - p(x)\right)^2 &=& q(x)\end{eqnarray}
expanding the left-hand side, we get:$$
q(x) = \left(2p(\frac x 2)^2+ 2q(\frac x 2) - 1 - p(x)\right)^2 + 16 p(\frac x 2)^2q(\frac x 2) \pm 8\left(2p(\frac x 2)^2+ 2q(\frac x 2) - 1 - p(x)\right)p(\frac x 2)\sqrt{q(\frac x 2)}
$$
which implies $\pm\sqrt{q(x/2)}$ is actually a rational function. Since its square is a polynomial, this means $\pm\sqrt{q(x/2)}$ is a polynomial itself, so $\pm\sqrt{q(x)}$ is also a polynomial. Therefore $\cos(x) = p(x) \pm \sqrt{q(x)}$ is a polynomial for $x\in[0,b-a]$. Since this interval contains the origin, we again have a contradiction, so $\cos(x)$ cannot be a polynomial on $[a,b]$.

As an addendum: All of these arguments can be generalized to show that $\cos(x)$ is also not a rational function on any interval, and that the other trig functions similarly are not polynomials or rational functions.