Proof:

Looking at the portion of the Mayer Vietoris sequence

$$\begin{CD} @> >> H_n(U\cap V)@> >> H_n(U)\oplus H_n(V) @> >> H_n(Y) @> >> H_{n-1}(U\cap V)@> >> \end{CD}$$

we get the result directly when $n\geq 2$. If $n = 1$ then necessarily $k\geq 3$ and we look at the portion of the Mayer Vietoris sequence

$$\begin{CD} @> >> H_1(U\cap V)@> >> H_1(U)\oplus H_1(V) @> >> H_1(Y) @> >> H_0(U\cap V)@>{\cong} >> H_0(U)\oplus H_0(V). \end{CD}$$

Observe that $H_1(U\cap V) = \{0\}$ and we get the exact sequence

$$0 \longrightarrow H_1(V)\longrightarrow H_1(Y) \longrightarrow 0,$$

establishing the result when $n = 1$. $\square$

The cases $n = k, k-1$ are more technical and we shall merely state the relevant results.