To ensure that OTP is a valid Shannon cipher, we check the decryption function.

$$\begin{array}{rl}\text{Dec}(k,\text{Enc}(k,m))& =\\ & =k\oplus (k\oplus m)\\ & =(k\oplus k)\oplus m\\ & =0\oplus m=m\end{array}$$

This indeed proves that decryption undoes encryption and so OTP is a valid private-key encryption scheme.

We claim that for every $m\in \mathcal{M}$, the distribution $D_m$ obtained by sampling the keyspace $k{\leftarrow }_R\mathcal{K}$ and outputting ${\text{Enc}}_k(m)$ is the uniform distribution over $\mathcal{C}$ and therefore, the distributions $D_m$ and $D_{m^{\prime }}$ are identical for every $m,m^{\prime }\in \mathcal{M}$.

Observe that every ciphertext $c\in \mathcal{C}$ is output by ${\text{Enc}}_k(m)$ if and only if $c=m\oplus k$. This in turn is true if and only if $k=m\oplus c$. The key $k$ is chosen uniformly at random from $\mathcal{K}$, so the probability that $k$ happens to be $m\oplus c$ is exactly $\frac{1}{|\mathcal{K}|}$. Moreover, the key, plaintext and ciphertext have the same length, so $|\mathcal{K}|=|\mathcal{M}|=|\mathcal{C}|$ which means that this probability is equal to $\frac{1}{|\mathcal{M}|}$, thus making the cipher perfectly secret.