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I'm attempting to calculate the de Rham cohomologies of the $n$-torus: $n \choose k$.

I'd like to use a Mayer-Vietoris sequence relating $H^kT^{n}$ to $H^kT^{n-1}$ and $H^{k-1}T^{n-1}$ so I can use the identity ${n \choose k} = {n-1 \choose k} + {n-1 \choose {k-1}}$ to achieve the result by induction. But I am having trouble choosing an open cover for the sequence. The obvious choice would be to split the torus along one of its circles, i.e. $T^n=T^{n-1} \times S^1=U \cup V$ with $U=T^{n-1} \times (1/4,3/4)$ and $V=T^{n-1} \times [0,1/2)\cup(1/2,1]$ where we are identifying the point $0$ and $1$ to make $[0,1]\cong S^1$.

Here's my beef: By retraction $H^k U \cong H^k V \cong H^k T^{n-1}$ and we'd have $$H^k(U \cap V)\cong H^k (T^{n-1}\sqcup T^{n-1})\cong H^k T^{n-1} \oplus H^k T^{n-1} \cong H^kU \oplus H^kV$$ so that the $T^{n-1}$ terms cancel each other out of the sequence and you just end up with $\sum_{k=0}^n (-1)^k \dim H^k T^n=0$ which is true but only helpful if I cared about the Euler characteristic. Am I mistaken? Is there a better cover for the sequence?

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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.

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