Lemma

Let X be a topological space connected and locally connected by arcs and let $x$ be a point of X. Suppose that the point $x$ has a countable fundamental system of neighbourhoods. Then, for every sequence $(x_n)_{n\geqslant 1}$ of points of X tending to $x$, there exists a path $c$ in X such that $c(0) =x$ and $c(1/n) =x_n$ for $n\geqslant 1$.

Let $(W_m)_{m\geqslant 1}$ be a fundamental system of neighbourhoods of $x$. Put $V_0= X$ and for every $m\geqslant 1$, let $V_m$ be an arc-connected neighbourhood of $x$ contained in $V_{m-1}\cap W_m$.

For every integer $n\geqslant 1$, denote by $m_n$ the largest integer $m\leqslant n$ such that $x_k\in V_m$ for every $k\geqslant n$. The sequence $(m_n)_{n\geqslant 1}$ is increasing by definition; it tends to infinity, because the sequence $(x_n)_{n\geqslant 1}$ tends to $x$. For every integer $n\geqslant 1$, let $c_n:\mathbf{I}\rightarrow V_{m_n}$ be a path with origin $x_{n+1}$ and endpoint $x_n$ in $V_{m_n}$. Define a mapping $c:\mathbf{I}\rightarrow X$ by putting $c(0) =x$ and $c(t) =c_n(n(n+ 1)t-n)$ if $1/(n+ 1)< t\leqslant$ $1/n$. We have $c(1/n) =x_n$ for every $n\geqslant 1$. The mapping $c$ is therefore continuous on every interval of the form $[1/(n+ 1),1/n]$ with $n\geqslant 1$, hence on the interval $]0,1]$. If $t\leqslant 1/n$, the point $c(t)$ belongs to $V_{m_n}$; the mapping $c$ is therefore continuous at 0.