If your manifold is closed you must be using some definition that ensures it's compact? So if $f : M \to \mathbb R^3$ is an embedding, isn't the image simultanously compact and open?
edit in response to your comment: if $f : M \to \mathbb R^3$ is an embedding, let $B \subset M$ be an open subset of $M$ which is homeomorphic to an open ball in $\mathbb R^3$. You need to argue that $f(B)$ is open in $\mathbb R^3$. If your embedding is smooth there's a big theorem from calculus that gives you the result. If you're talking about topological embeddings you're going to need a tool. Have you studied the "invariance of dimension" theorem?