Here are the statement and proof of the proposition, as they appear in the book:
Let $\mathbb{U}^2$ be the unit square $\{ x + iy : 0 \le x \le 1, \ 0 \le y \le 1\}$. [$\ldots$]
PROPOSITION 2.45. Let $\gamma_n : [0, 1] \to \mathbb{U}^2$ be a sequence of continuous paths, for $n \in \mathbb{N}$. Suppose that:
(i) The sequence of functions $(\gamma_n)$ is uniformly convergent.
(ii) The closure of the union of the images of all $\gamma_n$ is the whole of $\mathbb{U}^2$.
Then the limit $$ \gamma = \lim_{n\to\infty}\gamma_n $$ is continuous, and its image is $\mathbb{U}^2$. [$\ldots$]
Proof. By condition (i), a basic theorem in real analysis implies that the limit $\gamma$ exists and is continuous.
Let $z \in \mathbb{U}^2$. By condition (ii), for each $m > 0$, $m \in \mathbb{N}$, there exists $t_m \in [0, 1]$ and $n_m \in \mathbb{N}$ such that \begin{equation} \tag{1}\label{ineq:1} \left\lvert\gamma_{n_m}(t_m) - z\right\rvert < \frac{1}{m} \end{equation} The subsequence $\gamma_{n_m}$ also tends uniformly to $\gamma$. The sequence $(t_m)$ lies in $[0, 1]$ which is closed and bounded, so it has a convergent subsequence, with limit $t_0$. We claim that $\gamma(t_0) = z$. By uniform continuity, if $m$ is large enough, \begin{equation} \tag{2}\label{ineq:2} \left\lvert\gamma_{n_m}(t_0) - \gamma(t_0)\right\rvert < \frac{1}{m} \end{equation} So \begin{equation} \tag{3}\label{ineq:3} \left\lvert\gamma(t_0) - z\right\rvert < \frac{2}{m} \end{equation} for all $m$, so $\gamma(t_0) = z$ as claimed. $\square$
(Tags \eqref{ineq:1}, \eqref{ineq:2}, \eqref{ineq:3} have been added to make it easier to discuss the proof, and are not in the book.)
Assuming that the sequence $(n_m)_{m\geq1}$ was chosen to be strictly increasing, it is clear that $\gamma_{n_m}$ tends uniformly to $\gamma$, as stated.
The next statement is also clearly correct: there exists $t_0 \in [0, 1]$, and a strictly increasing sequence of positive integers $(m_k)_{k\geq1}$, such that the subsequence $u_k = t_{m_k}$ tends to $t_0$.
If we write (almost illegibly!) $\sigma_k = \gamma_{n_{m_k}}$, then $\sigma_k$ also tends uniformly to $\gamma$.
From \eqref{ineq:1}, we have: \begin{equation} \tag{4}\label{ineq:4} \left\lvert\sigma_k(u_k) - z\right\rvert = \big\lvert\gamma_{n_{m_k}}(t_{m_k}) - z\big\rvert < \frac{1}{m_k}. \end{equation}
Take any $\epsilon > 0$. By the continuity of $\gamma$ at $t_0$, there exists $\delta > 0$ such that: \begin{equation} \tag{5}\label{ineq:5} \left\lvert\gamma(t) - \gamma(t_0)\right\rvert < \frac{\epsilon}{3} \text{ if } \left\lvert t - t_0 \right\rvert < \delta. \end{equation}
Because $u_k \to t_0$ as $k \to \infty$, there exists $K_1$ such that: \begin{equation} \tag{6}\label{ineq:6} \left\lvert u_k - t_0 \right\rvert < \delta \text{ if } k > K_1. \end{equation}
By \eqref{ineq:5} and \eqref{ineq:6}, \begin{equation} \tag{7}\label{ineq:7} \left\lvert\gamma(u_k) - \gamma(t_0)\right\rvert <\frac{\epsilon}{3} \text{ if } k > K_1. \end{equation}
Because $\sigma_k \to \gamma$ uniformly as $k \to \infty$, there exists $K_2$ such that: \begin{equation*} \left\lvert\sigma_k(t) - \gamma(t)\right\rvert < \frac{\epsilon}{3} \text{ for all } t \in [0, 1] \text{ if } k > K_2. \end{equation*}
In particular: \begin{equation} \tag{8}\label{ineq:8} \left\lvert\sigma_k(u_k) - \gamma(u_k)\right\rvert < \frac{\epsilon}{3} \text{ if } k > K_2. \end{equation}
By \eqref{ineq:4}, because $m_k \to \infty$ as $k \to \infty$, there exists $K_3$ such that: \begin{equation} \tag{9}\label{ineq:9} \left\lvert\sigma_k(u_k) - z\right\rvert < \frac{\epsilon}{3} \text{ if } k > K_3. \end{equation}
By \eqref{ineq:7}, \eqref{ineq:8} and \eqref{ineq:9}, taking any $k > \max\{K_1, K_2, K_3\}$, \begin{equation*} \left\lvert\gamma(t_0) - z\right\rvert < \epsilon. \end{equation*}
Because $\epsilon$ was arbitrary, it follows that $\gamma(t_0) = z$, as claimed. $\square$
This proof could be accused of being tiresomely pedantic, and maybe even juvenile! It is certainly longer and more cluttered with notation than the proof in the book; but at least I think I can see that it is valid. (Is it, though?)
In contrast, although I have opened the book several times in the last few days, and stared intently at the proof, I can make no sense of its last two sentences (containing \eqref{ineq:2} and \eqref{ineq:3}). I seem to see at least three logical holes in it, but I won't belabour the reader with what may be only my misinterpretations.
Is the proof in the book valid, as it stands? If so, a good answer to this question might be a set of annotations spelling out the points I have failed to understand (which I must imagine the authors took to be 'obvious').
If there is indeed something wrong with it, can a simple edit restore sense? If so, such a suggested edit would also be a good answer.
(Corrections or improvements to my proof, or alternative proofs, would also be welcome.)
In the light of Gio67's counterexample, consider adding a third condition to the proposition:
(iii) for every $n$, the image of $\gamma_n$ is nowhere dense.
In any topological space (not only a complete metric space, as in the Baire Category Theorem), the intersection of two dense open sets is dense; therefore the union of two nowhere dense sets is nowhere dense; therefore the union of a finite number of nowhere dense sets is nowhere dense. Therefore, for every $m \geq 2$, the union $\Gamma$ of the images of $\gamma_1, \gamma_2, \ldots, \gamma_{n_{m-1}}$ is nowhere dense. It follows that we can choose $w \in \mathbb{U}^2$ such that $|w - z| < 1/m$ and $w \notin \Gamma$. Because $\Gamma$ is closed, $w$ has a neighbourhood that is disjoint from $\Gamma$ and contained in the open ball with centre $z$ and radius $1/m$. By condition (ii), we can find $\gamma_{n_m}(t_m)$ in this neighbourhood of $w$, and \eqref{ineq:1} is satisfied, with $n_m > n_{m-1}$, as required. $\square$
