I have started to learning about nets. I need a solution verification or any necessary comments (answer if needed) for the following problem:
Statement: Let $X$ be a topological space and let $f:X\to \mathbb{R}$ be continuous. Show the following:
$(1)$ If a net $\mathscr{S}:D\to X$ converges to $x_0$, then $f\circ \mathscr{S}$ converges to $f(x_0)$ ($(D, \geq)$ denote a directed set).
$(2)$ If $(f\circ \mathscr{S})(d)\leq 1$ for every $d\in D,$ show that $f(x_0) \leq 1.$
proof:
$(1)$ Consider an open interval $(a,b)$ in $\mathbb{R}$ containing $f(x_0)$. Obviously, $f^{-1}((a,b))$ is open in $X$. Since $\mathscr{S}$ converges to $x_0$, there exists $m\in D$ such that $\mathscr{S}(d)\in f^{-1}((a,b))$ for all $d\geq m$. Note that $f\circ \mathscr{S}:D\to \mathbb{R}$ is also a net in $\mathbb{R}$. And $f\circ \mathscr{S}(d)\in (a,b)$ for all $d\geq m$. Since the open interval was chosen arbitrarily, the proof follows.
$(2)$ If possible, suppose $f(x_0)>1$. Find disjoint pair of open sets $U,~V$ in $\mathbb{R}$ such that $f(x_0)\in U$, each member of $U$ is greater than $1$, and $1\in V$. Now, $f^{-1}(U)$ is open in $X$ containing $x_0$ since $\mathscr{S}$ converges to $x_0$, there exists $m\in D$ such that $\mathscr{S}(d)\in f^{-1}(U)$ for all $d\geq m$. Then $f\circ \mathscr{S}\cap U\neq \emptyset.$ This is contradiction since $f\circ \mathscr{S}(d) \leq 1$ for every $d\in D$.
Any help will be appreciated. Thanks in advance.
$(1)$ is fine.
For $(2)$ it's slightly easier: suppose $f(x_0) > 1$. Then $U := (1,\infty)$ is an open set of $\Bbb R$ that contains $f(x_0)$. Now apply $(1)$ so there exists $m \in D$ such that for all $d \ge m$ we have $f(\mathscr{S}(d)) \in U$ so $f(\mathscr{S}(d)) > 1$. The same contradiction ensues. Introducing $V$ too is unnecessary.