Apart from the given norm $\|\cdot\|$, we can define another norm by $$\|\sum_{j=1}^k \lambda_je_j\|_{\infty}=\max_{1\le j\le k} |\lambda_j|$$ where the $\lambda_j$ are scalars. (You should check that this is a well-defined norm.)
Any two norms on a finite-dimensional vector space are equivalent, so in particular there is a constant $C>0$ so that $\|x\|_{\infty}\le C\|x\|$ for all vectors $x$. This inequality shows that if $\|v_k-V\|\to 0$, then we must have $\|v_k-V\|_{\infty}\to 0$ as $k\to\infty$, which implies that each component of $v_k-V$ converges to $0$, or that each component of $v_k$ converges to the corresponding component of $V$.
Not sure if this is the fastest answer, but it gets the job done. Recall that a linear function $f$ between Banach spaces is continuous if and only if it is bounded, e.g. $\|f(x)\|\leq C_f\|x\|$.
As suggested in the comments, begin with
\begin{align*}
\|v_n(u_n(x))-v(u(x))\|_G &\le\|v_n(u_n(x))-v_n(u(x))\|_G + \|v_n(u(x))-v(u(x))\|_G \\
&\le\|v_n(u_n(x)-u(x))\|_G + \|v_n(u(x))-v(u(x))\|_G \\
\end{align*}
Since each $v_n$ is bounded, the first term is at most $C_{v_n}\|u_n(x)-u(x)\|_F$.
Let us take supremums on both sides and assume that we can remove the $n$-dependence from the constant $C_{v_n}$; we will justify this assumption at the end of the post. We then obtain:
$$\sup_{\|x\|_E\leq 1}\|v_n(u_n(x))-v(u(x))\|_G ~~\le~~ C\sup_{\|x\|_E\leq 1}\|u_n(x)-u(x)\| + \sup_{\|x\|_E\leq 1}\|v_n(u(x))-v(u(x))\|_G.$$
For sufficiently large $n$, the first term may be made strictly less than $\varepsilon/2$, directly from the convergence $u_n\to u$. The second term may also be made at most $\varepsilon/2$ as follows: since $\|x\|_E\leq 1$ we have $\|u(x)\|_F\leq C_u$. Thus, for sufficiently large $n$,
$$\sup_{\|x\|_E\leq 1}\|v_n(u(x))-v(u(x))\|_G \leq \sup_{\|y\|_F\leq C_u}\|v_n(y)-v(y)\|_G = C_u \sup_{\|y\|_F\leq 1}\|v_n(y)-v(y)\|_G<\varepsilon/2.$$
Hence, we have bounded $\|v_n\circ u_n - v\circ u\|$ by $\varepsilon$ for any sufficiently large $n$, as desired.
The "technical details" song-and-dance:
By taking the relevant supremum, we can remove the $n$-dependence in this constant as follows: since $v_n\to v$, we have that $\sup_{\|y\|_F\leq 1}\|v_n(y)-v(y)\|_F\leq 1$ for any $K$ and for all sufficiently large $n$, so that
$$\sup_{\|y\|_F\leq 1}\|v_n(y)\|\leq \sup_{\|y\|_F\leq 1}\|v_n(y)-v(y)\|_F + \sup_{\|y\|_F\leq 1}\|v(y)\|_F$$
and thus $C_{v_n}=\|v_n\|\leq 1+\|v\|$ for all sufficiently large $n$.
Moreover, since $u_n\to u$, we have that $\sup_{\|x\|_E\leq 1}\|u_n(x)-u(x)\|_F\leq 1$ for all sufficiently large $n$. Since the supremum displayed above therefore includes $y=u_n(x)-u(x)$, we have that for sufficiently large $n$,
$$\sup_{\|x\|_E\leq 1}\|v_n(u_n(x)-u(x))\|_G \leq (C_v+1)\sup_{\|x\|_E\leq 1}\|u_n(x)-u(x)\|_F.$$
Notice that $C=C_v+1$ is independent of $n$, as desired.
Best Answer
Suppose that $T$ is discontinuous at some $x_0\in X$. Then, since $T$ is linear, it is also discontinuous at $0$. So, there's some sequence $(v_n)_{n\in\Bbb N}$ such that $\lim_{n\to\infty}v_n=0$ and that you don't have $\lim_{n\to\infty}Tv_n=0$. So, for some $\varepsilon>0$, there are infinitely many $n$'s such that $\|Tv_n\|\geqslant\varepsilon$. You can assume without loss of generality that this occurs for every $n$. But then $\lim_{n\to\infty}\frac{v_n}{\sqrt{\|v_n\|}}=0$ too, $$\left\|T\frac{v_n}{\sqrt{\|v_n\|}}\right\|\geqslant\frac\varepsilon{\sqrt{\|v_n\|}}$$and, since $\lim_{n\to\infty}\sqrt{\|v_n\|}=0$,$$\lim_{n\to\infty}\left\|T\frac{v_n}{\sqrt{\|v_n\|}}\right\|=\infty.$$