Fix $\varepsilon>0$. We can find $N(\varepsilon)$ such that for all $n,m\geq N(\varepsilon)$ and all $x,y\in X$ with $x\neq y$: $\frac{|(f_n-f_m)(x)-(f_n-f_m)(y)|}{d(x,y)}\leq \varepsilon$. Since $f_n$ converges to $f$ pointwise, we have, taking the limit $m\to\infty$ for $x,y$ fixed that
\begin{equation}
(*) \quad \forall n\geq N(\varepsilon),\forall x,y\in X,x\neq y\quad\frac{|(f_n-f)(x)-(f_n-f)(y)|}{d(x,y)}\leq \varepsilon.
\end{equation} In particular for $n=N(\varepsilon)$ we have $\frac{|f(x)-f(y)|}{d(x,y)}\leq \varepsilon+\frac{|f_{N(\varepsilon)}(x)-f_{N(\varepsilon)}(y)|}{d(x,y)}$, which shows that $f$ is Lipschitz. Now we have to show that $|f-f_n|_L$ converges to $0$. But this is a consequence of $(*)$, since we only have to take the supremum over these $x$ and $y$.
Note that we don't need to use Ascoli's theorem to show that $\{f_n\}$ has a limit (just use the same proof when you show that the set of the continuous real functions on a compact space for the uniform norm is a Banach space).
Here compactness is needed to be sure that each Lipschitz function is bounded, since such a function is continuous.
L. Schwartz and A. Grothendieck made clear, by very early 1950s, that the Cauchy (-Goursat) theory of holomorphic functions of a single complex variable extended with essentially no change to functions with values in a locally convex, quasi-complete topological vector space. Cauchy integral formulas, residues, Laurent expansions, etc., all succeed (with trivial modifications occasionally).
Conceivably one needs a little care about the notion of "integral". The Gelfand-Pettis "weak" integral suffices, but/and a Bochner version of "strong" integral is also available.
Further, in great generality, as Grothendieck made clear, "weak holomorphy" (that is, $\lambda\circ f$ holomorphic for all (continuous) linear functionals $\lambda$ on the TVS) implies ("strong") holomorphy (i.e., of the TVS-valued $f$).
(Several aspects of this, and supporting matter, are on-line at http://www.math.umn.edu/~garrett/m/fun/Notes/09_vv_holo.pdf and other notes nearby on http://www.math.umn.edu/~garrett/m/fun/)
Edit: in response to @Christopher A. Wong's further question... I've not made much of a survey of recent texts to see whether holomorphic TVS-valued functions are much discussed, but I would suspect that the main mention occurs in the setting of resolvents of operators on Hilbert and Banach spaces, abstracted just a little in abstract discussions of $C^*$ algebras. (Rudin's "Functional Analysis" mentions weak integrals and weak/strong holomorphy and then doesn't use them much, for example.) Schwartz' original book did treat such things, and was the implied context for the first volume of the Gelfand-Graev-etal "Generalized Functions". In the latter, the examples are very small and tangible, but (to my taste) tremendously illuminating about families of distributions.
Edit-edit: @barto's further question is about the behavior of holomorphic operator-valued $f(z)$ at an isolated point $z_o$ where $f(z)$ fails to be invertible. I do not claim to have a definitive answer to this, but only to suggest that the answer may be complicated, since already for the case $f(z)=(T-z)^{-1}$ for bounded, self-adjoint $T$, it seems to take a bit of work (the spectral theorem) to prove that isolated singularities are in the discrete/point spectrum of $T$. But this may be overkill, anyway...
Best Answer
It is a complete metric space under the $L^{1}$ metric: $d(f,g)=\int|f-g|$. It cannot be a Banach space because it is not a vector space. Proof of the fact that this space is complete: if $f_n$ is a density function for each $n$ and $\{f_n\}$ is Cauchy then there exists $f \in L^{1}$ such that $f_n \to f$ in $L^{1}$ (because $L^{1}$ is complete). There is a subsequence which converges almost everywhere to $f$. Hence $f \geq 0$ almost everywhere. Of course, $\int f =\lim \int f_n$ so $\int f=1$. Hence $f$ is also a density function proving completeness.