de Rham comologies of the $n$-torus

Your covering is fine. Let's take a closer look at the maps defining the Mayer-Vietoris sequence: $$H^{k-1}(T^{n-1})\oplus H^{k-1}(T^{n-1})\xrightarrow{f}H^{k-1}(T^{n-1}\sqcup T^{n-1})\xrightarrow{\partial}H^k(T^n)\xrightarrow{g}H^k(T^{n-1})\oplus H^k(T^{n-1})$$

Although $f$ is a map between two isomorphic spaces, it is not an isomorphism, so the terms do not cancel. Let $([\alpha],[\beta])\in H^{k-1}(T^{n-1})\oplus H^{k-1}(T^{n-1})$. Then we have

$$f(([\alpha],[\beta]))=[\alpha-\beta]$$

I'm being a bit sloppy here with notation, using $\alpha$ instead of $i^*[\alpha]$. Similarly, for $[\alpha]\in H^k(T^n)$, we have $$g([\alpha])=([\alpha],[\alpha])$$ again with shorthand notation.

Finally, for any $[\alpha]\in H^{k-1}(T^{n-1}\sqcup T^{n-1})$, we may write $[\alpha]=[\beta]+[\gamma]$ reflecting the direct sum decomposition into disjoint components. Then $$\partial([\alpha])=[d\beta]=-[d\gamma]$$ where $d$ is the exterior derivative.

Now that we've had a closer look at the maps in the sequence, see if you can use the exactness of Mayer-Vietoris to show that there is a short exact seqence: $$0\rightarrow H^{k-1}(T^{n-1})\rightarrow H^{k}(T^n)\rightarrow H^k(T^{n-1})\to 0$$

From which your result will follow.