is any subset of a manifold a submanifold? by definition a submanifold is a subset of a manifold which is itself a manifold. consider $A$ a subset of an $n$-manifold $M$. a neighborhood of $x\in A$ is $\mathbb R^n$ since $x$ is an element of $M$ which is a manifold (all its elements have neighborhoods homeomorphic to  $\mathbb R^n$ does this mean that every subset $A$ of $M$ is a submanifold? my guess is no and i think there is some thing wrong with my notion of neighborhood but i can't see how?
 A: Whether $A$ is a manifold is determined by whether every point in $A$ has an $\mathbb R^k$-homeomorphic neighbourhood in $A$, not in $M$.
A: I can think of two ways of interpreting your question, for a subset S, of a manifold M:
1) Is S, with the subspace charts of M a manifold?
Answer: no, consider, e.g., f(x)=|x| , a standardly-embedded square.
But each of these is homeomorphic to a a manifold S' ; by smoothing the corners, or, more formally, pulling back manifold charts of S.
I think most people would use this interpretation. 
And:
2)Can the subset S of M be given any charts so that S is a manifold?
Edit 2: the previous example I used about a subset (not given the subspace topology)
of a manifold not being a manifold, may, at best, not be a good example. I think/hope
these cases of a subset S not given the subspace topology are better:
i) Take an uncountable subset S of $\mathbb R^n$ and give it the discrete topology.
Then S is not second countable, and we cannot modify S by a homeomorphism, into making
it a submanifold, unlike was the case of , e.g., a standardly-embedded square. 
Edit1: to answer you question, and to address the key point made by Theo, S as a
subset
of M is a submanifold if the subspace charts $\{(f_i,U_i)\}$ satisfy $f(U)$=
$(x_1,..,x_k,0,0,..0)$, so that $f_i, U_i$ must be/act as the injection of $\mathbb R^n$ 
into $\mathbb R^k$ 
