Derivation of standard error of mean

The only thing we need to prove here is that for any scalar constant $c$, and for a random variable $X$, $$\mathrm{Var}[cX] = c^2 \mathrm{Var}[X].$$ This follows from the property of expectation $$\mathrm{E}[cX] = c\mathrm{E}[X]$$ as follows: $$\begin{align*} \mathrm{Var}[cX] &= \mathrm{E}[(cX - \mathrm{E}[cX])^2] \\ &= \mathrm{E}[(cX - c\mathrm{E}[X])^2] \\ &= \mathrm{E}[c^2(X - \mathrm{E}[X])^2] \\ &= c^2 \mathrm{E}[(X - \mathrm{E}[X])^2] \\ &= c^2 \mathrm{Var}[X]. \end{align*}$$

Let $Y$ be any random variable. Let $Z = Y/n$. Then $$Z^2 = \frac1{n^2} Y^2,$$ $$E(Z^2) = E\left(\frac1{n^2} Y^2\right) = \frac1{n^2} E(Y^2)$$ and therefore $$E\left(\left(\frac Yn\right)^2\right) = \frac1{n^2} E(Y^2).$$ Also, $$E(Z) = E\left(\frac1n Y\right) = \frac1n E(Y).$$ So from $Var(Y)=E(Y^2)-(E(Y))^2$ and $Var(Z)=E(Z^2)-(E(Z))^2,$ we find $$\begin{eqnarray} Var\left(\frac Yn\right) = Var(Z) &=& E(Z^2)-(E(Z))^2\\ &=& \frac1{n^2} E(Y^2) - \left(\frac1n E(Y)\right)^2 \\ &=& \frac1{n^2} \left(E(Y^2) - \left( E(Y)\right)^2 \right) \\ &=& \frac1{n^2} Var(Y). \end{eqnarray}$$ Now consider the case where $Y = T$.