In the text General Topology of Stephen Willard is ask to show that if $(x_\lambda)_{\lambda\in\Lambda}$ is a net such that every its subnet has a subset converging to $x$ then $(x_\lambda)_{\lambda\in\Lambda}$ converges to $x$.
So to prove the statement I tried to work with a particular subset so that given the neighborhoods system $\mathcal U(x)$ of $x$ then I observed that the relation $$ (\lambda_1,U_1)\preccurlyeq(\lambda_2,U_2)\longleftrightarrow \lambda_1\preccurlyeq\lambda_2\wedge U_2\subseteq U_1 $$ for any $(\lambda_1,U_1),(\lambda_2,U_2)\in\Lambda\times\mathcal U(x)$ makes $\Lambda\times\mathcal U(x)$ a directed set. Now is not complicate to show that the projection $\pi_\Lambda$ onto $\Lambda$ of $\Lambda\times\mathcal U(x)$ is an increasing cofinal map so that the position $$ x_{(\lambda,U)}:=x_\lambda $$ for any $\lambda\in\Lambda$ and for any $U\in\mathcal U(x)$ defines a subnet of $(x_\lambda)_{\lambda\in\Lambda}$. So we observe that if $x_{(\lambda,U)}\to x$ then for any $U\in\mathcal U(x)$ there exists $(\lambda_0,U_0)\in\Lambda\times\mathcal U(x)$ such that $$ x_{(\lambda,U)}\in U $$ for any $(\lambda,U)\succcurlyeq(\lambda_0, U_0)$ but $(\lambda, U_0)\succcurlyeq(\lambda_0,U_0)$ for any $\lambda\succcurlyeq\lambda_0$ so that $$ x_\lambda=x_{(\lambda,U_0)}\in U $$ for any $\lambda\succcurlyeq\lambda_0$ which means that $x_\lambda\to x$. So I thought that I could prove the statement proving that if a subnet of $(x_{(\lambda, U)})_{(\lambda,U)\in\Lambda\times\mathcal U(x)}$ converges to $x$ then also $(x_{(\lambda, U)})_{(\lambda,U)\in\Lambda\times\mathcal U(x)}$ converges to $x$ but unfortunately I was not able to do this.
So how solve the exercise? Could someone help me, please?
So if $(x_\lambda)_{\lambda\in\Lambda}$ does not converge to $x$ then there exists $U\in\mathcal U(x)$ such that for any $\lambda\in\Lambda$ there exists $\nu_\lambda\in\Lambda$ such that $\nu_\lambda\succcurlyeq\lambda$ and such that $x_{\nu_\lambda}$ not lies in $U$ so that the set $$ \text{N}:=\{\nu\in\Lambda:x_\nu\notin U\} $$ is not empty. Now for any $\nu_1,\nu_2\in\text{N}$ there exists $\nu\in\Lambda$ such that $$ \nu\succcurlyeq\nu_1,\nu_2 $$ and thus what initially observed ensures that there exist $\lambda_\nu\in\Lambda$ such that $$ \lambda_\nu\succcurlyeq\nu\quad\text{and}\quad x_{\lambda\nu}\notin U $$ but this proves that $\lambda_\nu\in\text{N}$ and so we conclude that $\text{N}$ is a directed set: that is, we conclude that the restriction of the preorder $\preccurlyeq$ on $\Lambda$ makes $\text N$ a directed set. Now we observe that the inclusion map $$ \iota:\text{N}\ni\nu\to\nu\in\Lambda $$ is trivially increasing and moreover it is cofinal too because what initially observed trivially implies that for any $\lambda\in\Lambda$ there exists $\nu_\lambda\in\text{N}$ such that $\nu_\lambda\succcurlyeq\lambda$. So through the position $$ x_\nu:=x_{\iota(\nu)} $$ for any $\nu\in\text{N}$ we cand define a subnet $(x_\nu)_{\nu\in\text N}$ of $(x_\lambda)_{\lambda\in\Lambda}$ whose elements are disjoint from $U$ so that $x$ cannot be a cluster point for this subnet. However by assumption there exists a subnet of $(x_\nu)_{\nu\in\text N}$ converging to $x$ so that any subnet can be converge to $x$, but this is impossible since no of its elements is contained in $U$. So, for these reasons, we finally conclude that $(x_\lambda)_{\lambda\in\Lambda}$ converges to $x$.