Showing $\lbrace (x,y) \in \mathbb{R}^2:xy=1 \rbrace$ is Closed 
Let $K=\lbrace (x,y) \in \mathbb{R}^2:xy=1 \rbrace \subseteq \mathbb{R}^2$. Show that $K$ is closed.

I am following Munkres' topology book, and this is a step towards finishing problem 3 on p. 145. I'm doing a reading course, and my instructor gave me a hint about showing that the projection map is not a closed map, which lead me to this set.
If I know that the function $f: \mathbb{R}^2 \rightarrow \mathbb{R}$ defined by $f(x,y)=xy$ is continuous, then I am done, but proving that $f$ is continuous seems like more trouble than necessary, so I'm hoping to do it with a more elementary approach.
Of course I could do it by definition, but $K^c$ does not seem particularly easy to prove openness for. Limit points seem equally annoying to work with for this set, at least for me. I'm not even sure what limit points of $K$ would look like.
I've read this question: Showing a set is closed in $\mathbb R^2$
But most of the solutions, including the one given by the asker, use some results from analysis that I'm afraid I can't use.
Anyway, I'd like to know which of the "standard" ways to prove closedness seems appropriate to you, and a nudge in the right direction. Please no full solutions!
 A: Let $\{(x_n,y_n)\}$ be a sequence in $K$ with a limit $(x,y)\in\mathbb R^2$. Then $x_ny_n=1$ for all $n$, so $xy=1$. Hence $(x,y)\in K$ and $K$ is closed.
To @user46944's point, let $X$ be a topological space and $A\subset X$. Suppose $A$ contains all of its limit points. Then if $x\in A^c$, there exists a neighborhood $U$ of $x$ that does not intersect $A$. It follows that $A$ is closed. Conversely, suppose $A$ is closed. Let $x$ be a limit point of $A$. Suppose $x\in A^c$, then as $A^c$ is open, there exists a neighborhood $U$ of $x$ that does not intersect $A$. This contradicts the assumption that $x$ is a limit point of $A$.
A: The function $f: \mathbb R^2 \times \mathbb R^2 \to \mathbb R$ defined as $f(x,y) = xy$ is continuous because the projections $$\begin{align}\pi_1 : R^2 \times \mathbb R^2 &\to \mathbb R \\(x,y) &\mapsto x\end{align} $$
and $$\begin{align}\pi_2 : R^2 \times \mathbb R^2 &\to \mathbb R \\(x,y) &\mapsto y\end{align}$$
are continuous. This fact is easily showed by observing that both are Lipschitz, 
$$\|\pi_1 (x,y) - \pi_1(x',y') \| = \|x - x'\| \leq \|(x,y) - (x,y)\|$$  
valid for any norm taken in $\mathbb R^2$. And the product of two continuous functions is continuous. 
As $K$ is the pre-image of $\{1\} \subset \mathbb R$ by $f$  the result follows.  
A: If I were to prove this, these are the steps I would take (if you decide to try them and have any trouble, please comment and let me know):
Step 1.  First, show if $f: \Bbb R \to \Bbb R$ is continuous, then its graph (i.e., the set $K = \{ (x, f(x)) \mid x \in \Bbb R \}$) is a closed subset of $\Bbb R^{2}$ (hint: Use the $\epsilon-\delta$ definition of continuity to show the complement of $K$ is open).
Step 2. Then note that the $K$ described in your question is nothing more than $K = \{ (x, \dfrac{1}{x}) \mid x \in \Bbb R - \{ 0 \} \}$.  We know $f(x) = \dfrac{1}{x}$ is continuous on its domain $\Bbb R - \{0 \}$, so by the first argument you gave, its graph is closed as a subset of $\Bbb R^{2}$.
A: Here is a proof idea that relies a bit more on calculus. You should be able to picture the function $y = \frac{1}{x}$ in your mind, which can help you picture what I will say in the rest of this answer. Now let $(a,b) \in \mathbb{R}^2$ be an arbitrary point in the plane such that $ab \neq 1$. We can consider the shortest distance from $(a,b)$ to a point on the curve $y = \frac{1}{x}$. This distance will be determined by drawing a line from $(a,b)$ that is perpendicular to $y$ (you can use the derivative of $y$ to help solve for this perpendicular line and the length of it). You may get two lines projecting away from $(a,b)$ and hitting $y$ at a perpendicular angle. Just take the minimum of these two distances (let's say $d$), cut that distance in half and set $\frac{1}{2}d = \varepsilon$. Now you know the open ball $$B_\varepsilon((a,b)) = \{(c,d)\in \mathbb{R}^2: (a-c)^2+(b-d)^2<\varepsilon\}$$ is an open set that contains $(a,b)$, and by construction it will not intersect $y$. Since $(a,b)$ was arbitrary, we know any such coordinate in $\mathbb{R}^2$ where $ab \neq 1$ can be separated from $y$ with an open set, and hence cannot be a limit point of $y$. Then there are no limit points of $K$ not contained in $K$, which is to say if $K$ has a limit point, it is contained in $K$.
