(Update. Added a self-contained proof that under $\mathsf{ZF}+\mathsf{DC}+\mathsf{BP}$ there is no norm.)
Martín-Blas Pérez Pinilla is on the right track. You can't even put a norm on $C(\mathbb{R})$ without using the axiom of choice in an essential way. Dependent choice is not enough.
Claim. It is consistent with $\mathsf{ZF}+\mathsf{DC}$ that there does not exist any norm on $C(\mathbb{R})$.
Recall that a subset $E$ of a topological space is said to have the Baire property if it can be written as a symmetric difference $E = U \triangle M$ where $U$ is open and $M$ is meager (a countable union of nowhere dense sets).
A celebrated theorem of Shelah says that consistent with $\mathsf{ZF}+\mathsf{DC}$ is the statement $\mathsf{BP}$: "Every subset of $\mathbb{R}$ has the Baire property." From $\mathsf{BP}$ it follows that in fact every subset of any Polish space has the Baire property.
Let $\tau$ be the usual topology on $C(\mathbb{R})$ (uniform convergence on compact sets). It is induced by a translation-invariant metric:
$$d(f,g) := \sum_{n=1}^\infty 2^{-n} \min\left(1, \sup_{[-n,n]} |f-g|\right) $$
The metric $d$ is complete (this comes from the fact that a uniform limit of continuous functions is continuous). And it's not hard to see that $\tau$ is separable (the polynomials with rational coefficients are $\tau$-dense, by the Weierstrass approximation theorem). So $(C(\mathbb{R}),\tau)$ is a Polish space.
Suppose now that $\|\cdot\|$ is a norm on $C(\mathbb{R})$. Let $B$ be the closed unit $\|\cdot\|$-ball. We will show that $B$ does not have the Baire property with respect to $\tau$. Specifically, let $U$ be any $\tau$-open set; we will show that $B \triangle U$ is $\tau$-nonmeager.
Suppose first that $U = \emptyset$ so that $B \triangle U = B$. Since $\bigcup_{n=1}^\infty nB = C(\mathbb{R})$, by the Baire category theorem $B$ is $\tau$-nonmeager.
Now suppose that $U$ is nonempty. We will show $U \setminus B$ is $\tau$-nonmeager. Let us begin by showing $U \setminus B$ is nonempty. Let $f \in U$ and let $g \in C(\mathbb{R})$ be your favorite nonzero continuous function which is supported in $[0,1]$. Let $g_n(x) = g(x-n)$ be translates of $g$. For each $n$, let $a_n$ be a real number sufficiently large that $\|f + a_n g_n\| > 1$, so that $f + a_n g_n \notin B$. Then $a_n g_n \to 0$ uniformly on compact sets (i.e. in the $\tau$ topology), so for some $N$ we have $h := f + a_N g_N \in U$. Thus $h \in U \setminus B$. (To say this another way, every nonempty $\tau$-open set is unbounded, but $B$ is bounded, so $U$ cannot be a subset of $B$.)
Now for any $u \in C(\mathbb{R})$, we have $h + \frac{1}{n} u \to h$ in both the $\tau$ and $\|\cdot\|$ topologies. So for sufficiently large $n$, we have $h + \frac{1}{n} u \in U$ and $h + \frac{1}{n} u \in B^c$ (since $B$ is $\|\cdot\|$-closed). That is, $u \in n((U \setminus B)-h)$. Since $u$ was arbitrary we have shown $\bigcup_{n=1}^\infty n((U \setminus B)-h) = C(\mathbb{R})$. By the Baire category theorem, $U \setminus B$ is $\tau$-nonmeager, hence so is $B \triangle U$.
So we have shown that if $C(\mathbb{R})$ has a norm, then it has a set that lacks the Baire property with respect to $\tau$. Under $\mathsf{ZF}+\mathsf{DC}+\mathsf{BP}$ there is no such set and hence no norm.
(Credit where credit is due: This proof is loosely based on the idea of the proof of the Garnir-Wright closed graph theorem from Theorem 27.45 of Eric Schechter's Handbook of Analysis and its Foundations.)
Incidentally, the only property of the vector space $C(\mathbb{R})$ we used is that it admits a Polish topology in which every nonempty open set is unbounded. So the same argument would apply to other vector spaces with this property, such as $\mathbb{R}^{\mathbb{N}}$, $C^\infty(\mathbb{R}^d)$, etc.
