We know that pointwise convergence is a necessary condition of uniform convergence. Is this also a necessary condition in general if we consider another norm different from $\Vert\cdot\Vert_{\infty}$? More precisely:
I am wondering if it is possible to produce a norm $\Vert \cdot\Vert$ and a subset $X$ of the set of functions on some domain $M$ such that we find sequences of functions that don't converge pointwisely but converge with regard to $\Vert \cdot\Vert$.
Let $\mathbb{K}$ be your scalar field and $\mathbb{K}^M$ will denote the space of $\mathbb{K}$-valued functions on $M$.
For $m\in M$, define $\delta_m:\mathbb{K}^M\to\mathbb{K}$ by $\delta_m(f)=f(m)$. This is a linear functional on $\mathbb{K}$. If I'm understanding you correctly (that you want the norm to be only on a subspace $X$), what you're asking is equivalent to whether there exists a subspace $X$ and a norm $\|\cdot\|$ on $X$ such that for at least one $m\in M$, $\delta_a:(X,\|\cdot\|)\to \mathbb{K}$ is not continuous.
If $M$ is finite, then there is no such norm. Every subspace of $\mathbb{K}^M$ is finite-dimensional, and finite dimensional spaces don't admit discontinuous linear functionals.
On the other hand, if $M$ is infinite, start with $\ell_\infty(M)$, the space of bounded, $\mathbb{K}$-valued functions on $M$. Fix an arbitrary $m\in M$. Let $e_m\in \ell_\infty(M)$ be the function such that $e_m(m)=1$ and $e_m(n)=0$ for all $m\neq n \in M$. Let $$T=\{f\in \ell_\infty(m):\delta_m(f)=0\}.$$ Let $(f_i)_{i\in I}$ be a basis for $T$ and let $(f_i^*)_{i\in I}$ be the coordinate functionals. That is, each $f\in T$ can be written as a finite sum $f=\sum_{i\in I}a_if_i$ for $(a_i)_{i\in I}\in \mathbb{K}^I$, and $f^*_j(f)=a_j$.
At most finitely many of the functionals $(f_i^*)_{i\in I}$ can be continuous on $T$. Indeed, suppose that $\{i_1,i_2,\ldots\}$ is an infinite subset of $I$ such that each $f^*_{i_n}$ is continuous. Define $g_N=\sum_{n=1}^N\frac{f_{i_n}}{2^n\|f_{i_n}\|}\in T$ and $g=\sum_{n=1}^\infty \frac{f_{i_n}}{2^n\|f_{i_n}\|}\in T$. The series converges uniformly because it is absolutely convergent. It converges to a member of $T$ because $T$ is closed in $\ell_\infty(M)$. This means that $g_N\to g$. Because each $f_{i_n}^*$ is assumed to be continuous, $f_{i_n}^*(g)=\lim_N f^*_{i_n}(g_N)=\frac{1}{2^n\|f_{i_n}\|}$. But this is impossible, because, by the definition of a basis, $\{i\in I:f^*_i(g)\neq 0\}$ must be finite.
Choose $j\in I$ such that $f^*_j$ is discontinuous on $T$.
Any member of $\ell_\infty(M)$ can be written uniquely as $$h=h(m)e_m+f^*_j(h)f_j+\sum_{j\neq i\in I}f^*_i(h)f_i.$$ Define $$Ah=f^*_j(h)e_m+h(m)f_j+\sum_{j\neq i\in I}f^*_i(h)f_i.$$ Note that $A$ is an invertible linear operator (in fact, $A^2=A$). Define a new norm $|\cdot|$ by $$|h|=\|A\|_\infty,$$ where $\|\cdot\|_\infty$ is the usual norm. Then $(\ell_\infty(M),|\cdot|)$ is isometrically isomorphic to $(\ell_\infty(M), \|\cdot\|)$ (via the linear operator $A$), and is therefore complete. Moreover, $\delta_m$ is discontinuous on $(\ell_\infty(M),|\cdot|)$, because if it were continuous, $$f_j^*=\delta_m\circ A:(\ell_\infty(M),\|\cdot\|)\to (\ell_\infty(M),|\cdot|)\to \mathbb{K}$$ would be continuous, which it is not.
I will note that in order to find a basis for $T$, we used the Axiom of Choice. I believe we need something like the Axiom of Choice in order to guarantee the existence of any discontinuous linear functional on any Banach space.
We found one $m$ such that $\delta_m$ is discontinuous. You can probably modify what I did to get a norm such that many $\delta_m$s are discontinuous.