[proofplan] We exhibit the [discrete metric](/page/Discrete%20Metric) and compare its metric topology with the given topology. The key observation is that every metric ball of radius $1/2$ around a point is exactly the singleton containing that point. Hence every subset is open in the metric topology, while every metric-[open set](/page/Open%20Set) is automatically open in the given discrete topology because the latter contains all subsets of $X$. [/proofplan] [step:Define the discrete metric on $X$] Define the map $d:X\times X\to\mathbb{R}$ by declaring that $d(x,y)=0$ when $x=y$ and $d(x,y)=1$ when $x\neq y$. This is precisely the discrete metric on $X$. Since $X$ is a set and $d$ is the discrete metric on $X$, [citetheorem:8358] implies that $(X,d)$ is a [metric space](/page/Metric%20Space). Let $\tau_d$ denote the metric topology on $X$ induced by $d$. [/step] [step:Compute the small metric balls] For each $x\in X$, define the open $d$-ball of radius $1/2$ centered at $x$ by \begin{align*} B_d(x,1/2)=\{y\in X:d(x,y)<1/2\}. \end{align*} We claim that $B_d(x,1/2)=\{x\}$ for every $x\in X$. Indeed, if $y\in B_d(x,1/2)$, then $d(x,y)<1/2$. Since $d(x,y)$ can only be $0$ or $1$, this forces $d(x,y)=0$, and by the definition of the discrete metric this means $y=x$. Thus $B_d(x,1/2)\subset\{x\}$. Conversely, $d(x,x)=0<1/2$, so $x\in B_d(x,1/2)$ and therefore $\{x\}\subset B_d(x,1/2)$. Hence $B_d(x,1/2)=\{x\}$. [guided] Fix a point $x\in X$. The metric topology is generated by open balls, so we look for a ball that isolates $x$ from every other point. Define \begin{align*} B_d(x,1/2)=\{y\in X:d(x,y)<1/2\}. \end{align*} The choice of radius $1/2$ is useful because the discrete metric only takes the two values $0$ and $1$. If $y\in B_d(x,1/2)$, then $d(x,y)<1/2$. The value $1$ is not less than $1/2$, so the only possible value is $d(x,y)=0$. By the definition of the discrete metric, $d(x,y)=0$ exactly when $x=y$. Therefore every point of $B_d(x,1/2)$ is equal to $x$, giving $B_d(x,1/2)\subset\{x\}$. For the reverse inclusion, the center $x$ belongs to its own ball because $d(x,x)=0<1/2$. Hence $\{x\}\subset B_d(x,1/2)$. Combining the two inclusions gives \begin{align*} B_d(x,1/2)=\{x\}. \end{align*} [/guided] [/step] [step:Show every subset is open in the metric topology] Let $A\subset X$. For each $x\in A$, the previous step gives $\{x\}=B_d(x,1/2)$, so $\{x\}\in\tau_d$. Since $\tau_d$ is closed under arbitrary unions, \begin{align*} A=\bigcup_{x\in A}\{x\} \end{align*} belongs to $\tau_d$. This also covers the case $A=\varnothing$, since the empty union is open in every topology. Therefore $\mathcal{P}(X)\subset\tau_d$. [/step] [step:Compare the metric topology with the discrete topology] Every element of $\tau_d$ is a subset of $X$, so $\tau_d\subset\mathcal{P}(X)$. Since the given topology is discrete, $\tau=\mathcal{P}(X)$. The previous step gives $\mathcal{P}(X)\subset\tau_d$, and hence \begin{align*} \tau_d=\mathcal{P}(X)=\tau. \end{align*} Thus the metric $d$ induces the topology $\tau$. [/step] [step:Conclude metrizability] We have exhibited a metric $d:X\times X\to\mathbb{R}$ such that the metric topology $\tau_d$ equals the given topology $\tau$. By the definition of metrizability, $(X,\tau)$ is metrizable. [/step]