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For a topological space $X$ (not finite)

Is there a way to construct a new space $Y$ such that $X \subsetneq Y$ (i.e., $\exists \ i:X \rightarrow Y$, an embedding not surjective) and $X \simeq Y$ (homeomorphic)?

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    If $X \subset Y$ means the set $X$ is contained in the set $Y$ then the answer is yes. If $X \subset Y$ means that $Y$ contains a homeomorphic copy of $X$ then the answer is no. Please specify. – 2017-02-27
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    @Daron Can you elaborate on why there might not be a space $Y$? What's an example of a space $X$ such that no homotopy equivalent space contains it as a homeomorphic copy? – 2017-02-27

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No, not in general. For instance, if $X$ has only finitely many points, then if $X\subsetneq Y$ then $Y$ must have more points than $X$, so $X$ cannot be homeomorphic to $Y$.

For an infinite counterexample, consider $X=S^1$. If $X\subsetneq Y$ and $X\simeq Y$, that means that there is a proper subset of $X$ that is homeomorphic to $X$ (namely, the preimage of $X\subset Y$ under a homeomorphism $X\to Y$). But no such subset exists, since every connected proper subset of $X$ is an interval.

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    What about in the infinite case. It's pretty obvious that the question has an affirmative "no" when considering the case (as you have shown). – 2017-02-27
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    @Eric Well, you are right. But I mean when X is not finite. I'll change the hypothesis. – 2017-02-27
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No. Consider the circle. Assume $f: S^1 \to S^1$ is an embedding. Then its degree has to be $\pm 1$ (by a local argument), and therefore it's surjective.