A problem on the convergence of a sequence I am reading "Calculus vol.1" which was written by Matsusaburo Fujiwara and published in 1934 in Japan.
At first I thought I could prove the following theorem in the above book easily, but I could not prove that because I am a beginner.

If $u_n^{(k)} \xrightarrow{n\to\infty} u^{(k)}$, $u^{(k)} \xrightarrow{k\to\infty} u_0$ and if you choose $n_1, n_2, \ldots$ appropriately, $u_{n_k}^{(k)} \xrightarrow{k\to\infty} u_0$ holds.

Please tell me the proof of the above theorem.
 A: I'm assuming $\big(u_n^{(k)}\big)_{n=1}^\infty$ are real sequences. 
Let $\varepsilon > 0$.
Since $u_n^{(1)} \xrightarrow{n\to\infty} u^{(1)}$ there exists $n_1 \in \mathbb{N}$ be such that $n \ge n_1 \implies \left|u^{(1)}-u^{(1)}_{n}\right| < \frac{\varepsilon}{2}$.
Since $u_n^{(2)} \xrightarrow{n\to\infty} u^{(2)}$ there exists $n_1 \in \mathbb{N}$, $n_2 \ge n_1$ such that $n \ge n_2 \implies \left|u^{(2)}-u^{(2)}_{n}\right| < \frac{\varepsilon}{4}$.
Continuing this iteratively, we obtain an increasing sequence $(n_k)_{k=1}^\infty$ such that $n \ge n_k \implies \left|u^{(k)}-u^{(k)}_{n}\right| < \frac{\varepsilon}{2^k}$.
Since $u^{(k)} \xrightarrow{k\to\infty} u_0$, there exists $k_0 \in\mathbb{N}$ such that $k\ge k_0 \implies \left|u^{(k)} - u_0\right| < \frac\varepsilon2$.
Now, for $k \ge k_0$ we have:
$$\left|u^{(k)}_{n_k} - u_0\right| \le \left|u^{(k)}_{n_k} - u^{(k)}\right|+\left|u^{(k)} - u_0\right| < \frac{\varepsilon}{2^k}+\frac{\varepsilon}{2} < \varepsilon$$
Therefore, we have $u^{(k)}_{n_k} \xrightarrow{k\to\infty} u_0$.
A: Let $m\in\Bbb N$. Take a number $n_m$ s.t. $|u_{n_m}^{(k)}(x)-u^{(k)}(x)|\le\frac{1}{m}$ for any $x$. Then use the triangle inequality starting from the diference $|u_{n_m}^{(k)}(x)-u_0(x)|$. Pass with $m$ to $\infty$.
A: I imitated Mr. mechanodroid's solution.
I'm assuming $\big(u_n^{(k)}\big)_{n=1}^\infty$ are real sequences. 
Let $\varepsilon > 0$.
Since $u_n^{(k)} \xrightarrow{n\to\infty} u^{(k)}$, there exists $n_k \in \mathbb{N}$ such that $\left|u^{(k)}_{n_k}-u^{(k)}\right| < \frac{\varepsilon}{2}$.
Since $u^{(k)} \xrightarrow{k\to\infty} u_0$, there exists $k_0 \in\mathbb{N}$ such that $k\ge k_0 \implies \left|u^{(k)} - u_0\right| < \frac\varepsilon2$.
Now, for $k \ge k_0$ we have:
$$\left|u^{(k)}_{n_k} - u_0\right| \le \left|u^{(k)}_{n_k} - u^{(k)}\right|+\left|u^{(k)} - u_0\right| < \frac{\varepsilon}{2}+\frac{\varepsilon}{2} = \varepsilon$$
Therefore, we have $u^{(k)}_{n_k} \xrightarrow{k\to\infty} u_0$.
