For any bounded subset $E$ of the real line, if f is continuous over the entire real line, is $f(E)$ also bounded?

I understand that if I take the Euclidean metric space and some bounded subset $E$ of the real line $\mathbb{R}$ that $f(E)$ is not necessarily bounded if $f: E \to \mathbb{R}$ is continuous.

However, what if my domain is $\mathbb{R}$, so my function $f:\mathbb{R} \to \mathbb{R}$ is continuous over the entire real line. For any bounded subset $E$ of the real line, is $f(E)$ necessarily bounded too?

If $f$ is continuous on $\mathbb{R}$, then $f$ is continuous on the closure $\bar{E}$ of $E$, which is compact. Hence $f(\bar{E})$ is bounded, so $f(E)$ is too.

Yes, it is true. Suppose that $f$ is not bounded on $E$. WLOG assume that $f$ takes arbitrarily large values. Pick a sequence of values $v_1,v_2,...$ with limit $+\infty$. Suppose for every $i$, $v_i=f(u_i)$, $u_i\in E$. Since the sequence $(u_i)$ is bounded it has a subsequence $u_{j_k}$ which has a limit $u$. Since $f$ is continuous, $+\infty = \lim v_{j_k}=\lim_{k\to\infty} f(u_{j_k})=f(u)$, a contradiction.