Proving infinite wedge sum of circles isn't first countable

Let $\{S_i\}_{i=1}^\infty$ be a countable family of circles and $\{p_i\}_{i=1}^\infty$ be a family of points such that $p_i\in S_i$. let $X = \bigcup _{i=1}^\infty S_i/\{p_i\}_{i=1}^\infty$ be the topological space obtained from the quotient map $q: \bigcup _{i=1}^\infty S_i\longrightarrow X$ by collapsing all $p_i$ to one point $p\in X$. Show that $X$ is not first countable.

I had this diagonalization idea: suppose for the sake of contradiction that there's a countable neighbourhood basis $\{U_i\}_{i=1}^\infty$ at point $p\in X$. let $V_{ij}=q^{-1}(U_i)\cap S_j$.

At this point my intuition suggests that $W=\bigcup_{i=1}^\infty V_{ii}$ is a saturated open set that satisfies $$\forall i>0(U_i\nsubseteq q(W))$$ thus arriving at a contradiction but all my attempts so far to prove it rigorously have failed.

Any help formalizing my argument will be appreciated.

Solution 1:

HINT: For each $k\in\Bbb Z^+$ there is an $r_k>0$ such that

$$B(p_k,r_k)\subseteq q^{-1}[U_k]\;,$$

where $B(x,\epsilon)$ is the usual open ball. For $k\in\Bbb Z^+$ let $\epsilon_k=\frac12r_k$, and consider the image under $q$ of the saturated open set

$$\bigcup_{k\in\Bbb Z^+}B(p_k,\epsilon_k)\;.$$