= 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 \b[$Y$ 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 $x_n\not\to x$, there are $\varepsilon>0$ and a subsequence with $d(x_{n_j},x)\geq\varepsilon$. <Sequential compactness> supplies a further subsequence converging to $x'$. By <continuity> of $f$, $f(x')=\lim y_{n_j}=y=f(x)$. Injectivity implies $x'=x$, contradicting the fixed lower distance bound. Hence $x_n\to x$. Sequential <continuity> is equivalent to <continuity> in <metric spaces>, proving \b[$f^{-1}$ 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 $\operatorname{Re}z>0$, the normalization $z\mapsto z/|z|$ is continuous and takes values in the open right semicircle. Therefore
$$
\boxed{\operatorname{Arg}(z)=g^{-1}(z/|z|)\in(-\pi/2,\pi/2)}
$$
is a continuous argument satisfying $z=|z|e^{i\operatorname{Arg}z}$. Equivalently it is $\arctan(\operatorname{Im}z/\operatorname{Re}z)$.
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