# Let $X \subseteq \mathbb R$ and $X$ has same cardinality as $\mathbb R$ , does there always exist a continuous surjection from $\mathbb R$ onto $X$ ?

Let $X \subseteq \mathbb R$ and $X$ has same cardinality as $\mathbb R$ , does there always exist a continuous surjection from $\mathbb R$ onto $X$ ? ( I know that there need not always be a continuous surjection from $X$ onto $\mathbb R$ , for example when $X$ is closed bounded interval )

No. For instance, take $X = \mathbb R \setminus \mathbb Q$, the irrationals. Then $X$ is uncountable and totally disconnected, i.e. its connected components are single points. Since $\mathbb R$ is connected, any continuous map $f: \mathbb R \to X$ sends $\mathbb R$ into one connected component of $X$. Thus $f(\mathbb R)$ is a single point, hence $f$ cannot be surjective.
In fact, this works for any disconnected subset of $\mathbb R$ which is uncountable. Just take $X = \mathbb R \setminus \{0\}$, for instance.
Not even close. There are $2^{\aleph_0}$ continuous functions from $\Bbb R$ to itself, so there are only $2^{\aleph_0}$ continuous images of $\Bbb R$.
However there are $2^{2^{\aleph_0}}$ subsets of $\Bbb R$ which have the same cardinality as $\Bbb R$ itself. So by cardinality argument alone we see that most sets cannot be obtained in such way.
Indeed, the easy argument as given by Alex G. in the other answer is that a continuous image of $\Bbb R$ must be connected, so any disconnected subset cannot be obtained like that. But what if we replace $\Bbb R$ by a Borel set instead? There are totally disconnected Borel sets, like the irrationals or Cantor sets.