This question seems to be posed too quickly (without substantial thinking) and has an (almost trivial) affirmative answer: We should prove that $\mathrm{Cont}(\mathbb R,\mathbb R)$ intersects each non-empty open set $U\subset \mathbb R^{\mathbb R}$. We can assume that $U$ is of basic form: $U=\prod_{r\in\mathbb R}U_r$, where for each $r\in\mathbb R$ the set $U_r$ is open in $\mathbb R$ and the set $F=\{r\in\mathbb R:U_r\ne\mathbb R\}$ is finite. Now take any (piecewise linear) continuous function $f:\mathbb R\to\mathbb R$ such that $f(r)\in U_r$ for any $r\in F$. Then $f\in \mathrm{Cont}(\mathbb R,\mathbb R)\cap U$. So, $\mathrm{Cont}(\mathbb R,\mathbb R)$ is dense in $\mathbb R^{\mathbb R}$ (even for the Tychonoff product topology of the real lines endowed with the discrete topology).

This question seems to be posed too quickly (without substantial preliminary thinking) and has an (almost trivial) affirmative answer: We should prove that $\mathrm{Cont}(\mathbb R,\mathbb R)$ intersects each non-empty open set $U\subset \mathbb R^{\mathbb R}$. We can assume that $U$ is of basic form: $U=\prod_{r\in\mathbb R}U_r$, where for each $r\in\mathbb R$ the set $U_r$ is open in $\mathbb R$ and the set $F=\{r\in\mathbb R:U_r\ne\mathbb R\}$ is finite. Now take any (piecewise linear) continuous function $f:\mathbb R\to\mathbb R$ such that $f(r)\in U_r$ for any $r\in F$. Then $f\in \mathrm{Cont}(\mathbb R,\mathbb R)\cap U$. So, $\mathrm{Cont}(\mathbb R,\mathbb R)$ is dense in $\mathbb R^{\mathbb R}$ (even for the Tychonoff product topology of the real lines endowed with the discrete topology).

This question seems to be posed too quickly (without substantial thinking) and has an (almost trivial) affirmative answer: We should prove that $\mathrm{Cont}(\mathbb R,\mathbb R)$ intersects each non-empty open set $U\subset \mathbb R^{\mathbb R}$. We can assume that $U$ is of basic form: $U=\prod_{r\in\mathbb R}U_r$, where for each $r\in\mathbb R$ the set $U_r$ is open in $\mathbb R$ and the set $F=\{r\in\mathbb R:U_r\ne\mathbb R\}$ is finite. Now take any (piecewise linear) continuous function $f:\mathbb R\to\mathbb R$ such that $f(r)\in U_r$ for any $r\in F$. Then $f\in \mathrm{Cont}(\mathbb R,\mathbb R)\cap U$. So, $\mathrm{Cont}(\mathbb R,\mathbb R)$ is dense in $\mathbb R^{\mathbb R}$.

This question seems to be posed too quickly (without substantial thinking) and has an (almost trivial) affirmative answer: We should prove that $\mathrm{Cont}(\mathbb R,\mathbb R)$ intersects each non-empty open set $U\subset \mathbb R^{\mathbb R}$. We can assume that $U$ is of basic form: $U=\prod_{r\in\mathbb R}U_r$, where for each $r\in\mathbb R$ the set $U_r$ is open in $\mathbb R$ and the set $F=\{r\in\mathbb R:U_r\ne\mathbb R\}$ is finite. Now take any (piecewise linear) continuous function $f:\mathbb R\to\mathbb R$ such that $f(r)\in U_r$ for any $r\in F$. Then $f\in \mathrm{Cont}(\mathbb R,\mathbb R)\cap U$. So, $\mathrm{Cont}(\mathbb R,\mathbb R)$ is dense in $\mathbb R^{\mathbb R}$ (even for the Tychonoff product topology of the real lines endowed with the discrete topology).