Fix a ground field $k$. It is well-known that for an infinite dimensional vector space $V$ the canonical map $V \to V^{**}$ is not an isomorphism since cardinalities do not coincide ($V^{**}$ being much more bigger than $V$). I've heard that one can solve this issue by introducing a certain topology on $V$, namely by letting a collection $(U_{\alpha})_{\alpha}$ of vector subspaces to be an open local base at zero and letting the topology be translation invariant. Moreover, if one endows $k$ with the discrete topology $V$ turns out to be a topological vector space.
How is the topology in $V^{*}$ (now only considering continuous linear functionals) defined such that the canonical map $V \to V^{**}$ is an isomorphism?
I believe I must be missing some hypotheses on $(U_\alpha)_\alpha$, maybe the resultant topology needs to be separated or even complete in the sense $$ \varprojlim_\alpha V/U_\alpha \cong V $$ Any comment or reference is appreciated.
This is not true in general. For instance, if $k=\mathbb{R}$ and you take $V=\mathbb{R}$ with its usual topology (not the discrete topology), then $V^*$ is trivial and hence $V^{**}$ is trivial as well for the only possible topology on $V^*$.
What is true is that the canonical map $V\to V^{**}$ is always surjective, for a certain natural topology on $V^*$. This topology on $V^*$ is the weak* topology, also known as the topology of pointwise convergence. That is, consider $V^*$ as a subspace of $k^V$ with the product topology. This topology does make the canonical map $V\to V^{**}$ well-defined, since evaluation at each element of $V$ is continuous on $V^*$ (and in fact $V^*$ has the coarsest topology which makes this true). Now suppose $f:V^*\to k$ is a continuous linear functional. Since $f$ is continuous, $\ker f$ is open, so it contains a basic open neighborhood of $0$ in the product topology. This means there exist finitely many $v_1,\dots,v_n\in V$ such that for all $\alpha\in V^*$ such that $\alpha(v_i)=0$ for all $i$, $\alpha\in\ker f$.
Let $V_0\subseteq V$ be the span of $v_1,\dots,v_n$; then for any $\alpha\in V^*$, $f(\alpha)$ depends only on the restriction of $\alpha$ to $V_0$. Let $W$ be the space of linear functionals on $V_0$ that extend to continuous linear functionals on $V$. Define $f_0:W\to k$ by $f_0(\alpha_0)=f(\alpha)$, for any $\alpha\in V^*$ extending $\alpha_0$. Then $f_0$ is a linear functional on $W$, and can be extended to a linear functional on the entire algebraic dual of $V_0$. Since $V_0$ is finite-dimensional, this implies there exists $v\in V_0$ such that $f_0(\alpha_0)=\alpha_0(v)$ for all $\alpha_0\in W$. It follows that $f(\alpha)=\alpha(v)$ for all $\alpha\in V^*$, so $f$ is the image of $v$ under the canonical map $V\to V^{**}$.
Note moreover that the kernel of $V\to V^{**}$ consists of those $v\in V$ which are annihilated by every element of $V^*$. So $V\to V^{**}$ is an isomorphism iff for each nonzero $v\in V$, there exists a continuous linear functional $V\to k$ which does not vanish on $v$. (This is a strong condition on the topology of $V$, and in particular it implies that every finite-dimensional subspace of $V$ has the discrete topology and is a topological direct summand of $V$. Any vector space does admit a topology with this property, though, e.g. the discrete topology.)