Here's an alternate proof which doesn't use invariance of domain. It also gives a slightly stronger result.
Theorem:
Let $M^n$ be compact without boundary. Then there is no immersion $f:M\rightarrow \mathbb{R}^n$.
Proof: (sketch). Assume for a contradiction there is such an $f$. Since $M$ is compact, $f$ is a closed map, that is, it maps closed sets to closed sets. To see this, let $F\subseteq M$ be closed. Then it's compact because $M$ is, so $f(F)$ is compact, hence closed. (Here, we just use the fact that $f$ is continuous).
Further, $f$ is an open map. That it, it maps open sets to open sets. To see this, note that it is enough to map really small open sets to open sets because $f(\bigcup U_i) = \bigcup f(U_i)$.
It's a fact (a consequence of the implicit function theorem, if I recall) that every immersion locally looks like the natural inclusion $\mathbb{R}^k\rightarrow \mathbb{R}^n$ into the first $k$ coordinates. (This uses the boundarylessness of $M$ - if $p$ is on the boundary, this part doesn't work.)
Said more precisely, given our immersion $f$ and a point $p\in M$, there is a chart around $p$ and around $f(p)$ so that in these chart coordinates, $f$ looks like the inclusion.
Now, in our case $k = n$ and then the natural inclusion is a homeomorphism. This implies open sets in these special charts are mapped to open sets, so $f$ is open.
Putting this together, we've now seen that $f$ is an open and closed mapping. Now, what is $f(M)$? It must be compact because $M$ is, but it must also be both open and closed in $R^n$ because $M$ is open and closed in itself. This implies $f(M) = \mathbb{R}^n$ since that's the only nonempty clopen subset of $\mathbb{R}^n$, but $\mathbb{R}^n$ isn't compact, so we've reached a contradiction.
Best Answer
There are two notions here: manifold, and manifold-with-boundary. If you add the boundary circle to a Möbius strip, you get a manifold-with-boundary, which is not a manifold. The statement that all compact manifolds embedded in $\mathbb R^3$ is true. For $0$ and $1$-manifolds, it follows because all $0$ and $1$-manifolds are orientable. For surfaces it follows because any connected compact surface in $\mathbb R^3$ divides $\mathbb R^3$ into two pieces, which takes a little work to prove. (I like Alexander duality as a proof.) Then you can orient the surface by taking an outward-pointing normal vector at each point and using the right-hand rule to orient the tangent-plane at each point.
The statement is false for $\mathbb R^4$. You can embed the projective plane in $\mathbb R^4$.