Continuous image of separable space

As pointed out in the comments, you may finish your argument by noticing that $f(\overline{A})\subset \overline{f(A)}$: if $y\in f(\overline{A})$, then $y=f(x)$ for $x\in\overline{A}$, thus if $V\subset Y$ is an open set such that $y\in V$, then $f^{-1}(V)\subset X$ is an open set (because $f$ is continuous) and $x\in f^{-1}(V)$, from which it follows that $f^{-1}(V)\cap A\ne\emptyset$, hence $V\cap f(A)\ne\emptyset$.

Alternatively, you may prove that $f(A)$ is dense by showing that $f(A)\cap V\ne\emptyset$ for all nonempty open sets $V\subset f(X)$. For if $\emptyset\ne V\subset f(X)$ is an open set, then $\emptyset\ne f^{-1}(V)\subset X$ is open, thus there exists $a\in A$ such that $a\in f^{-1}(V)$, showing that $f(a)\in V$.

Since $V$ is arbitrary, this shows that $f(A)$ is dense in $f(X)$.

With $Z=f(X) ,$ the function $f:X\to Z$ is a continuous surjection and we wish to how that $Z$ is separable.

With closure bars denoting closure in $X$ or in $Z,$ we have $$f(\overline A)\subset \overline {f(A)}\subset Z$$ because $f$ is continuous. We have $$Z=f(X)=f(\overline A)$$ because $X=\overline A.$ So we have $$Z=f(X)=f(\overline A)\subset \overline {f(A)}\subset Z.$$ Therefore $\overline {f(A)}=Z.$ So $f(A)$ is a countable dense subset of $Z.$