For $p \geq 1$ define a metric$$d'(x, y) = \left(\sum_{i=1}^{n} |x_i -y_i |^{p}\right)^{\frac{1}{p}}$$ for $x, y \in \mathbb{R}^n$ for $x, y \in \mathbb{R}^n$. Show this metric induces the usual topology on $\mathbb{R}^n$
My Attempted Proof
The topology induced by $d'$ has as its basis $\mathcal{B} = \{B_{d'}(x, \epsilon) \ | \ x \in \mathbb{R}^n, \epsilon > 0\}$, where $B_{d'}(x, \epsilon) = \{ y \ | \ d'(x, y) < \epsilon\}$.
Let $\mathcal{B}'$ denote the basis for the standard topology on $\mathbb{R}^n$. Take $x \in U \in \mathcal{B}'$, where $U = (a_i, b_i) \times ... \times (a_n, b_n)$. For each $i$, there exists a $\epsilon_i > 0$ such that $(x_i - \epsilon_i, x_i + \epsilon_i) \subset (a_i, b_i)$. Put $\epsilon = \min\{\epsilon_1, ..., \epsilon_n\}$, and by choosing a large enough $p$ we have $B_{d'}(x, \epsilon) \subset U$.
Conversely we take $y \in B_{d'}(x, \epsilon) \in \mathcal{B}$. Since each $U \in \mathcal{B}'$ is of the form $U = (a_i, b_i) \times ... \times (a_n, b_n)$, there exists a $U \in \mathcal{B}$ such that $U = (x_1 - \epsilon, x_1 + \epsilon) \times ... \times (x_n - \epsilon, x_n + \epsilon)$ and we have $y \in U \subset B_{d'}(x, \epsilon)$. $\square$
Is my proof correct? If so how rigorous is it? Also any comments on my proof writing (e.g where I can improve) and any criticism on my proof is greatly appreciated.