Consider a continuous function defined all of $R^n$ i.e., $h:R^n \to R$. Also obtain the infimum of $h$ $$ a: = \inf_{x\in R^n} h(x) $$
Define the set $A := \{ x \in R^n : h(x) > a \}$. Is $A$ open?
I was trying to show that it is open. If $A$ is all of $R^n$ then it is open. We will consider the case when $A \subset R^n$. If $A$ is a strict subset of $R^n$ then its complement exists. Denote it by $A^c$. Also, since $h(x)$ is defined over all $R^n$ and $a$ is the infimum, we can obtain $A^c$ explicitly as $$ A^c = \{ x \in R^n : h(x) = a \}$$
It remains to show that $A^c$ is closed. Suppose that $A^c$ is not closed. Then there is a sequence of points in $A^c$: $x_1, x_2, \dots$ such that $\lim_{n\to \infty} x_n \notin A^c$. This is a contradiction since $h$ is continuous and $ h(x_n) = a$ for every $n$, and its limit is $a$.
Did I get this right? Is there a better proof?
Your proof works. However, there is an easier way. Note that $B = \{y\in \Bbb R\mid y>a\}$ is open in $\Bbb R$, and $A = f^{-1}(B)$. Since $f$ is continuous, that means that $A$ is open. (This works even if $a = -\infty$, in which case $B = \Bbb R$ and $A = \Bbb R^n$.)