Proposition 1

Let $f$ be a mapping of a topological space $X$ into a topological space $X'$. If $f$ is continuous at $x$, and if $x$ lies in the closure of a subset $A$ of $X$, then $f(x)$ lies in the closure of $f(A)$.

Let $V'$ be a neighbourhood of $f(x)$ in $X'$. Since $f$ is continuous at $x$, $\overline{f}^{-1}(V')$ is a neighbourhood of $x$ in $X$. Hence $\overline{f}^{-1}(V')$ meets $A$, from which it follows that $V'$ meets $f(A)$, and therefore $f(x)$ is in the closure of $f(A)$.