Topology - multiplication of real valued functions $fg$ is continuous. Problem: If $f,g: X \rightarrow \mathbb{R}$ are continuous real valued functions on a topological space $X$, then so is $fg$. 

I could show that, by standard consequences of product topology, that the function $f\times g: X \times Y \rightarrow X' \times Y'$ is continuous, given continuous functions $f:X \rightarrow X'$ and $g:Y\rightarrow Y'$. 

Similarly, I could also show that the 'diagonal' map, $\Delta: X \rightarrow X \times X$ defined by $\Delta (x) = (x,x)$ is continuous. 

Now $fg = m \circ (f \times g) \circ \Delta$, where
$m: \mathbb{R} \times \mathbb{R} \rightarrow \mathbb{R}$ is the multiplication of real numbers. It suffices to show $m$ is continuous.

It seems to me that $m$ must be continuous, but I could not really understand how: if $U$ is open $\mathbb{R}$ then $m^{-1}(U)$ is open in $\mathbb{R} \times \mathbb{R}$. How should I approach?
Let $x$ be in $U$ then there is an $\varepsilon&gt;0$ such that $]x-\varepsilon, x+\varepsilon [\ \subset U$.

For $m((y,z))=x$ choose $\delta &lt; \frac{1}{3} \inf \left\{ \frac{\varepsilon}{y}, \frac{\varepsilon}{z}, \sqrt{\varepsilon}  \right\}$, then for $(y',z')\in \ ]y- \delta, y+\delta[\ \times \ ]z-\delta , z+\delta[$

$$ y'z' \in \  ]x-\varepsilon, x+\varepsilon [ .$$

This means any point $(y,z)$ in $m^{-1}(U)$ the open ball $]y- \delta, y+\delta[ \ \times\ ]z-\delta , z+\delta[$ is contained in $ m^{-1}(U)$.
Since $\mathbb{R}$ is metric space，how about using sequential criterion for continuity? That is $x_n\to x,y_n\to y$, we have $x_ny_n\to xy$.

Or use the definition of continuity in metric space, that is 
$\forall (x_1,y_1)\in \mathbb{R}\times \mathbb{R},\forall \epsilon &gt; 0,\exists \delta&gt;0$, such that for all $(x_2,y_2)\in \mathbb{R}\times \mathbb{R}$, with $d((x_1,y_1), (x_2,y_2)) &lt;\delta$, we have $|x_2y_2-x_1y_1|&lt;\epsilon$.