Is the following proof correct?
Let $(x_n)_n, (y_n)_n$ be two sequences which both converge to a limit $p$. i.e. $$(\forall \epsilon ∈ ℝ_{>0})(\exists N_{\epsilon, x}∈ ℕ)(\forall n ≥ N_{\epsilon, x})(d(p, x_n) < \epsilon)$$ and $$(\forall \epsilon ∈ ℝ_{>0})(\exists N_{\epsilon, y}∈ ℕ)(\forall n ≥ N_{\epsilon, y})(d(p, y_n) < \epsilon).$$
Consider the sequence $(z_n)_n$ given by $z_n = x_{2n}$ for $n ∈ 2ℤ$ and $z_n = y_{2n+1}$ for $n ∈ 1+2ℤ$. Now let $\epsilon ∈ ℝ_{>0}$ be arbitrarily given, and set $N_{\epsilon, z} := \max{\{N_{\epsilon/2, x}, N_{\epsilon/2, y}\}}$. Let $n ≥ N_{\epsilon, z}$. Now if $n ∈ 2ℤ$, then we have $$d(p, z_n) ≤ d(p, x_n) + d(x_n, z_n) = d(p, x_n) + d(x_n, x_{2n}) < \epsilon/2 + \epsilon/2? = \epsilon $$ Analogously for $n ∈ 2ℤ$.
Here I'm bluffing that $x_n$ and $x_{2n}$ are also close enough, but I don't believe this is actually true.. (For example, in the beginning of the sequence it does not need to be true, it seems.) However, it seems fixable. Should I use Cauchyness of $(x_n)_n$ to find another $N$ and include that in the maximum or something to make it true..?
You should state what you are trying to prove. I think you are trying to prove that any sequence that is made up of $\{x_n\}$ and $\{y_n\}$ interleaved together must also converge to p.
In that case, for any given $e$:
1) Take $N_1$ to be the index in $\{x_n\}$ where all terms after $N_1$ are within $e$ of $p$
2) Take $N_2$ to be the index in $\{y_n\}$ where all terms after $N_2$ are within $e$ of $p$
3) Take $N_1'$ to be the position in the interleaved sequence where $x_{N_1}$ is located
4) Take $N_2'$ to be the position in the interleaved sequence where $y_{N_2}$ is located
5) Take N to be $max(N_1', N_2')$. Then we are done.