Let $X$ and $J\neq\emptyset$ be sets s.t for every $j\in J$ $(Z_j,\tau _j)$ is a topological space. Let $\tau $ be weakest topology on $X$ s.t for every $j\in J$, $f_j : X\to Z_j$ is continous. Let $x\in X$ and $(x_\alpha)_{\alpha\in\mathcal A}$
Show that if for every $j\in J$, $f_j(x_\alpha)\to f_j(x)$, then $x_\alpha\to x$
By convergence, we mean $x_\alpha\to x\iff \forall U_x, \exists\alpha _0 : \forall\alpha (\alpha _0\preceq\alpha\Longrightarrow x_\alpha\in U_x)$, where $U_x$ is a neighborhood of $x$.
Initial thought: Assume for a contradiction, for some $U_x\in\tau $, for every $\alpha$, exists $\alpha _0$ s.t $\alpha\preceq\alpha _0$ and $x_{\alpha _0}\notin U_x$.
Due to continuity of $f_j,j\in J$ we have for every $V_{f_j(x)}\in\tau _j$ exists $\alpha _0$: $$\forall\alpha (\alpha _0\preceq\alpha \Longrightarrow f_j(x_\alpha)\in V_{f_j(x)}) $$ Therefore $x_\alpha\in f_j^{-1}(V_{f_j(x)})\in\tau $, trouble is that potentially any preimage of $V_{f_j(x)}$ for any $j\in J$ might "miss" the fixed $U_x$ (aside from $x$ itself, of course, but isn't contained in $U_x$). An idea $$\bigcup_{j\in J}\bigcup_{V\in\mathcal{V}_{f_j(x)}}f_j^{-1}(V)\overset{?}=\bigcup_{U\in\mathcal{U}_x}U $$ Where $\mathcal{U}_x$ is the collection of all $\tau$ open neighborhoods $x$ and analogous description for $\mathcal{V}_{f_j(x)}$.
It's obvious that $\subset$, but it's not so obvious that $\supset$ and it might not even be true.
What to do? I'm probably over-thinking this problem waaay out of proportion.
A basis of $\tau$ is given by the family of sets of the form
$$U = \bigcap_{j \in J_0} f_j^{-1}(W_j),$$
where $J_0$ is a finite subset of $J$, and for every $j \in J_0$, $W_j\in \tau_j$.
Thus, for $x\in X$, a neighbourhood basis of $x$ is given by the family of sets of the form
$$U = \bigcap_{j \in J_0} f_j^{-1}(V_j),\tag{$\ast$}$$
where $J_0$ is a finite subset of $J$ and $V_j$ is a neighbourhood of $f_j(x)$ for each $j \in J_0$.
Let's suppose we have a net $(x_{\alpha})_{\alpha \in \mathcal{A}}$ in $X$ such that $f_j(x_{\alpha}) \to f_j(x)$ for every $j$. We want to show that $x_{\alpha} \to x$.
So let $U$ be an arbitrary neighbourhood of $x$. Choose a neighbourhood $U'$ of $x$ of the form $(\ast)$ such that $U' \subset U$. By assumption, $f_j(x_{\alpha}) \to f_j(x)$ for every $j\in J_0$, so there is an $\alpha_j \in \mathcal{A}$ such that $\alpha_j \preceq \alpha$ implies $f_j(x_{\alpha}) \in V_j$. Since $\mathcal{A}$ is a directed set, there is an $\alpha_{J_0} \in \mathcal{A}$ such that $\alpha_j \preceq \alpha_{J_0}$ for every $j \in J_0$. But then
$$x_{\alpha} \in \bigcap_{j \in J_0} f_j^{-1}(V_j),$$
i.e. $x_{\alpha} \in U' \subset U$, for every $\alpha\in \mathcal{A}$ such that $\alpha_{J_0} \preceq \alpha$. Since $U$ was arbitrary, we have $x_{\alpha} \to x$.