Answer

First recall what the terms mean. Brownian motion is a stochastic process (set of random variables indexed by time) that starts at 0, has memoryless and Gaussian, 0-centered increments whose variance is the time difference. Mathematically, for any times $s \gt t$ and assuming level of information (filtration) $F_t$: $$B_0=0$$ $$B_s-B_t \sim N(0, s-t)$$ $$B_s-B_t \perp F_t$$

What about a martingale? Martingales are stochastic processes which are knowable at time $t$ (i.e. process $X_t$ is known / measurable at time t, no longer random), and best guess for their tomorrow's value given knowledge up to today, is today's value. Mathematically, assuming level of information (filtration) $F_t$ and times $s\gt t$: $$B_t \text{is adapted (measurable) t}F_t$$ $$E(B_s|F_t) = B_t$$Now knowing that $B_t$ is both a Brownian motion and a martingale, how can we show that $B_t^2-t$ is also a martingale? Well to show it's a martingale, we need to first see if it's measurable at $F_t$. Since $B_t$ is measurable at $F_t$ (it's first martingale property), then so is $B_t^2$ (if you know $B_t$ you also know $B_t^2$). And the time index $t$ is simply an index, so measurable too. First property is therefore satisfied: $B_t^2-t$ is adapted to $F_t$

For second property we must check that $E(B_s^2-s|F_t) = B_t^2-t$ for some times $s \gt t$. To do this, we'll have to somehow take advantage of the fact that $B_t$ is a brownian motion. All its properties are to do with increments though, so let's try to get an increment $B_s-B_t$ into our expression, by rewriting: $$E(B_s^2-s|F_t) = E((B_s-B_t+B_t)^2 - s + t - t|F_t)$$We've simply added and substracted $B_t$ and $t$. We can expand the square to:$$E((B_s-B_t)^2 + B_t^2 + 2B_t(B_s-B_t) -s + s -t |F_t)$$Since expectation is additive, we can further separate this to: $$E((B_s-B_t)^2|F_t) + E(B_t^2 -t|F_t) + 2E(B_t(B_s-B_t)|F_t) + E(t-s|F_t)$$Now we know the first term is expectation of a squared $N(0,s-t$ variable by second property of Brownian motion. Because the increment $B_s-B_t$ has expectation 0, the first term is simply equal to its variance (recall $Var(X) = E(X^2) - E(X)^2$ - in our case last term is 0). So first term is the variance of a $N(0, s-t$ variable, which is simply $s-t$.

As for the second term, we already proved that $B_t^2-t$ is measurable (known) at time $t$, so it's no longer a random variable. This means we can safely remove the conditional expectation and end up with $B_t^2-t$.

Fourt term: recall the third property of Brownian motion, where increment is independent of what came before it. That means $B_s - B_t$ is independent of anything up to time $t$, including $B_t$. Now we know that expectation of a product of independent random variables is product of their expectations, so we can write $$2E(B_t(B_s-B_t)|F_t) = 2E(B_t|F_t)E(B_s-B_t|F_t)$$We've already seen the increment $B_s - B_t$ has expectation 0, so this whole term is 0.

As for the last term, this is simply a difference of time indices, which are constants, so last term is simply $t-s$. Bringing all the terms together: $$E(B_s^2-s|F_t) = s-t + B_t^2 - t+ 0 + t-s = B_t^2-t$$Which is exactly what we needed to show to prove that $B_t^2-t$ is a martingale.