Let $X$ be a topological space. Show that the set $$C(X)=\{f:X \rightarrow \mathbb{R} : f \text{ is continuous}\}$$ is a linear subspace of all functions from $X$ to $\mathbb{R}$.
I am able to show that it is closed under multiplication but am unable to see how I could show the same under addition.
Let $f$ and $g$ be continuous functions from $X$ to $\mathbb R$. Consider $\{x\in X: f(x)+g(x) <c\}$ where $c$ is a real number. You can write this as $\bigcup_q \{x\in X: f(x)<q<g(x)-c\}$ where the union is over rational numbers $q$. Now write this sets as $\bigcup_q [f^{-1}(-\infty, q) \cap g^{-1} (q+c ,\infty)]$. It follows that $\{x\in X: f(x)+g(x) <c\}$ is open. Similarly $\{x\in X: f(x)+g(x) >c\}$ is open for any real number $c$. Taking intersection of sets of these two types you see that $(f+g)^{-1} (a,b)$ is open whenever $a<b$. Since any open set in $\mathbb R$ is a union of open intervals you see that inverse image of any open set under $f+g$ is open.