$\mathbb{R}\times\mathbb{R}$ $\cong$ $\mathbb{R}^2$ In my topology class we proved that $\mathbb{R}\times\mathbb{R}$ $\cong$ $\mathbb{R}^2$, when $\mathbb{R}$ and $\mathbb{R}^2$ has the metric topology and $\mathbb{R}\times\mathbb{R}$ has the standard topology. When we did the proof mye professor said that it was enough to show that if U$\times$V was open in $\mathbb{R}\times\mathbb{R}$ then U$\times$V is open in $\mathbb{R}^2$ as well. And then to show that if V was open in $\mathbb{R}^2$ then V was open in $\mathbb{R}\times\mathbb{R}$ as well. I do not understand how this is enough to show homeomrphism, how ever i'am guessing it might have something to do with the fact that a homeomorphism is a bijection where f(U) is open when U is open? Also could you prove this by defining f: $\mathbb{R}^2$ $\rightarrow$ $\mathbb{R}\times\mathbb{R}$, f(a,b)$\rightarrow$ a$\times$b? and then showed that f$\circ$ $f^{-1}$ is the identety map?

A topology on a space is defined by what the open sets are. So your professor's proof is enough because it shows that the open sets in $\mathbb{R}\times \mathbb{R}$ (with the product topology) are precisely the same as the open sets in $\mathbb{R}^2$ (with the metric topology). So the identity map $\mathbb{R} \times \mathbb{R} \to \mathbb{R}^2$ is a homeomorphism.

As Jason DeVito pointed out, showing $f \circ f^{-1}$ and $f^{-1}\circ f$ are identities does not prove homeomorphism unless you also prove that $f$ and $f^{-1}$ are continuous. Proving they are continuous consists of proving the two implications your professor has suggested.