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This fact is mentioned liberally in literature along with subsequent mention of Heine-Borel and I am trying to get my head around it. What would be a formal proof of this if we take for example, a unit sphere in $\mathbb R^3 $?

In addition, how could it be shown, non-trivially of course, that this sphere is not homeomorphic with $\mathbb R^2 $? As I understand this involves finding a bijective function with continuous inverse?

HeinousBarrel
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    a subset of $\mathbb{R}^n$ is compact iff it is closed and bounded. Compactness is a topological invariant and the plane is not compact. – yoyo May 05 '13 at 14:52
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    To show the sphere is not homeomorphic with the plane, it is enough to observe that the sphere is compact while the plane is unbounded and hence non-compact. – Sayantan May 05 '13 at 14:53
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    Many books on topology or analysis contain proofs of these fundamental results. See Topology by Munkres, or Principles of Mathematical Analysis by Rudin, for example. – Jim Belk May 05 '13 at 14:56

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As yoyo says in the comments, you can use the compactness criterium that a subset of $\mathbb{R}^n$ is compact if and only if it is closed and bounded.

So you can embed the sphere $S^2$ in $\mathbb{R}^3$ via the equation $x^2+y^2+z^2=1$. Satisfying an algebraic eqution is "clearly" a closed condition, so $S^2$ is closed (why is this clear? If you perturb the variables with arbitrarily small $\epsilon$, you will leave the sphere - this is what closedness means).

Why is it bounded? This is clear: the equation says precisely that any point on the sphere has absolute value $1$. So the sphere is clearly bounded.

It follows from Heine-Borel that the sphere is compact as a subset of $\mathbb{R}^3$. But compactness is a topological invariant, so we are done.

Fredrik Meyer
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    Another way to convince yourself ${ (x,y,z)\in\mathbb{R}^3 \mid x^2+y^2+z^2=1 }$ is a closed set, is to see that the map $f:\mathbb{R}^3 \to\mathbb{R}$ given by $f(x,y,z)=x^2+y^2+z^2$ is a continuous mapping. Then the inverse image under $f$ of a closed set is a closed set, and we are looking at $f^{-1}( { 1 } )$. – Jeppe Stig Nielsen May 27 '14 at 14:15