 Solution

Take any sequence $(y_n)$ in $Y$. Surjectivity lets us choose $x_n\in X$ with $f(x_n)=y_n$. By sequential compactness of $X$, some $x_{n_j}\to x\in X$. Continuity gives $y_{n_j}=f(x_{n_j})\to f(x)\in Y$. Thus $Y$ has the Bolzano–Weierstrass property.

Now suppose $f$ is a bijection. To prove continuity of its inverse, let $y_n\to y$ and put $x_n=f^{-1}(y_n)$, $x=f^{-1}(y)$. If $x_n\nrightarrow x$, there are $\epsilon >0$ and a subsequence with $d(x_{n_j},x)\ge \epsilon$. Sequential compactness supplies a further subsequence converging to $x^{\prime }$. By continuity of $f$, $f(x^{\prime })=\lim y_{n_j}=y=f(x)$. Injectivity implies $x^{\prime }=x$, contradicting the fixed lower distance bound. Hence $x_n\to x$. Sequential continuity is equivalent to continuity in metric spaces, proving $f^{-1}$ is continuous.

Apply this to $g(t)=e^{it}$ on $[-\pi /2,\pi /2]$, mapping onto the closed right semicircle $S$. This interval is sequentially compact, and $g$ is a continuous bijection: two equal exponential values differ by an integral multiple of $2\pi$, impossible for distinct points of an interval of length $\pi$. Consequently $g^{-1}:S\to [-\pi /2,\pi /2]$ is continuous. For $\mathrm{Re}z>0$, the normalization $z\mapsto z/|z|$ is continuous and takes values in the open right semicircle. Thereforeis a continuous argument satisfying $z=|z|e^{i\mathrm{Arg}z}$. Equivalently it is $\arctan (\mathrm{Im}z/\mathrm{Re}z)$.