Universal property of completion of a field at an absolute value 
Prove that the completion $\hat k$ of a field $k$ at one of its absolute values $|\ |$ satisfies the following universal property: every topological field embedding of $k$ into a complete field $k'$ extends uniquely to an embedding of $\hat k$ into $k'$ that is an isomorphism if and only if $k$ is dense in $k'$.

I think $(\implies)$ is easy.
I want to prove the other way implication. I define $\hat i : \hat k \rightarrow k'$ as $ \hat i([(x_n)])=\text {lim} \ (i(x_n))$.
The inverse of the above map can be defined as follows:
For any element, there is a sequence converging to that element in $k$. Send this element to equivalence class of that sequence.
As this is an isomorphism of topological fields I need to show $\hat i$ is a homeomorphism.
I am stuck here.    
I have made the following attempt. Please let me know imperfections or unwarranted assumptions etc.  
Assume that topology on $k'$ comes from an absolute value and embedding is continuous.
$$\hat i ^{-1}(B_{k'}(x, \epsilon))=\{ [(x_n)  ] \ | \ \text{lim} \ i(x_n) \in B_{k'}(x, \epsilon) \}   $$ $i$ being continuous, $y \in i^{-1}B_{k'}(x, \epsilon), \ \exists \delta \ : |z-y| < \delta \implies z\in  i^{-1}B_{k'}(x', \epsilon') \subset B_{k'}(x, \epsilon) $
So if $|[(x_n)]-[(y_n)]| < \delta  $ then after some $N$,  $|x_n-y_n| < \delta $ I am having difficulty proceeding hereafter.
 A: Recall that the topology of a field with an absolute value $|\cdot|$ is defined by the metric (a.k.a. distance) $d$ such that $d(x,y)=|x-y|$. This metric is translation invariant, i.e., $d(x+z, y+z)=d(x,y)$. Instead of fields with absolute values, let us investigate commutative groups with translation-invariant metric (NCP). We have the following proposition.
Lemma: A continuous group homomorphism between two NCPs maps cauchy sequences to cauchy sequences.
Proof: Let $i:(g, d)\to(g', d')$ be a continuous group homomorphism. Let $(x_n)$ be a cauchy sequence in $g$. 
Let $B_{g'}(0, \epsilon)$ be the open sphere in $g'$ of radius $\epsilon>0$ centered at $0$. Since $i^{-1}\left(B_{g'}(0, \epsilon)\right)$ is open in $g$, it contains $B_{g}(0, \beta(\epsilon))$ for some $\beta(\epsilon)\gt0$. We can assume $\beta$ is a non-increasing function since, if not, we can reset $\beta(\epsilon)=\min_{m\in\mathbb N, m\ge \frac1\epsilon}\beta(\frac 1m)$ for all $\epsilon$. 
Given $\epsilon>0$, since $(x_n)$ is a cauchy sequence, we can find $n_0=n_0(\epsilon)$ such that for all $n_1, n_2\ge n_0$, $d(x_{n_1}, x_{n_2})\lt\beta(\epsilon)$. Since $d(x_{n_1}, x_{n_2})=d(x_{n_1}-x_{n_2},0)$, $\ (x_{n_1}-x_{n_2})\in B_g(0,\beta(\epsilon))$. 
$$d'\left(i(x_{n_1}), i(x_{n_2})\right)= d'\left(i(x_{n_1})-i(x_{n_2}),0\right)= d'\left(i(x_{n_1}-x_{n_2}),0\right)\lt \epsilon, $$
which means $(i(x_n))$ is a cauchy sequence in $g'.$ $\blacksquare$.

Let us prove the following proposition.

Let $(g,d)$ be an NCP. Let $(\hat g, d)$ be the completion of $g$. Let $(g',d)$ be another NCP. (We abuse $d$ to mean the metric on difference spaces. Its meaning should be clear depending on the context.) $g'$ is complete. Let $i:g\to g'$ be a continuous group homomorphism. We have the following.
     (gI). There is a unique way to extend $i$ to a continuous map from $\hat g$ to $g'$.
     (gII). If $i$ is, in fact, a topological group embedding and $i(g)$ is dense in $g'$, that extension of $i$ must be a topological group isomorphism.

Proof for (gI). Let $[(x_n)]$ be an element in in $\hat g$, i.e., $x_n$ is a cauchy sequence in $g$. The lemma says $(i(x_n))$ is a cauchy sequence in $g'$. Since $g'$ is complete, $\lim\, (i(x_n))$ is well-defined. 
Suppose $\hat i : \hat g \to g'$ is a continuous map that extends $i$. Note that $x_1, x_2, \cdots$ converges to $[(x_n)]$ in $\hat g$. Since a continuous map between metric space must map convergent sequences to convergent sequences, $ \hat i([(x_n)])$ must be $\lim\, (i(x_n))$. On the other hand, we can verify that the map from $\hat g\to g'$ defined by $[(x_n)]\to\lim\, (i(x_n))$ is well-defined and continuous. So, we have proved there is a unique continuous map that extends $i$ to $\hat g$, which is $\hat i$ that is defined by $ \hat i([(x_n)])=\lim\, (i(x_n))$. $\blacksquare$.
Proof for (gII). In the proof for (gI), we have shown that the extension is $\hat i$ defined by $\hat i([(x_n)])=\lim\, (i(x_n))$ for $[(x_n)]\in \hat g$. 
It is routine to verify that $\hat i$ is a group morphism. 
Suppose $[(x_n)]$ and $[(y_n)]$ are two points in $\hat g$ such that $\hat i([(x_n)])=\hat i[(y_n)])\in g'$. Then $\hat i\left([(x_n-y_n)]\right)=\hat i\left([(x_n)]-[(y_n)]\right)=0$, which means $\lim\  (x_n-y_n) = 0$, i.e., $[(x_n)]$ and $[(y_n)]$ are the same points. So $\hat i$ is injective. $\hat i: \hat g\to \hat i(\hat g)$ is a group isomorphism. 
We have shown that $\hat i$ is bijective from $\hat g$ to $\hat i(\hat g)$. Since $\hat i$ is continuous, to show $\hat i: \hat g\to \hat i(\hat g)$ is a homeomorphism, we just need to show $\left(\hat i\right)^{-1}$ is continuous. 
Suppose $(i(\hat{y_n}))$ is a cauchy sequence in $i(\hat g)$, where $\hat{y_n}=[(y_{n,1}, y_{n,2}, \cdots)]\in \hat g$ for some $y_{n,m}\in g$. For each $n$, select $\beta(n)>0$ such that $d(y_{n,\beta(n)}, \hat{y_n})\lt \frac 1n$. Verify that $(\hat i(y_{n,\beta(n)}))$ is cauchy sequence in $i(\hat g)$. That means, $(i(y_{n,\beta(n)}))$ is a cauchy sequence in $i(g)$,
Note that the assumption $i: g\to i(g)$ is a homeomorphism implies that $i^{-1}:i(g)\to g$ maps cauchy sequences to cauchy sequences. So, $(y_{n,\beta(n)})=(i^{-1}(i(y_{n,\beta(n)}))$ is a cauchy sequence in $g$. Let $\hat y=[(y_{n,\beta(n)})]\in \hat g$. We can verify that that $\lim \hat{y_n} = \hat{y}$, i.e., $(\hat y_n)$ is a cauchy sequence. Since $(\hat y_n)=\left(i^{-1}(i(\hat y_n))\right)$, we have shown $\left(\hat i\right)^{-1}$ maps all cauchy sequences to cauchy sequences. Since both $\hat g$ and $\hat i(\hat g)$ are metric spaces, $\left(\hat i\right)^{-1}$ is continuous.
So $\hat i: \hat g\to \hat i(\hat g)$ is a homeomorphism and, hence, a topological group isomorphism.
Since $\hat i$ is a topological group isomorphism between metric spaces, the fact that $\hat g$ is complete implies $\hat i(\hat g)$ is complete. Since $\hat i(\hat g)\supseteq i(g)$ is dense in $g'$, $\hat i(\hat g)=g'$. So $\hat i$ is a topological group isomorphism. $\blacksquare$.

The proposition (gI) and (gII) can be applied to the additive groups of (complete) fields with absolute values. It becomes easy and routine to prove the following clearer and stronger version of the direction "$\impliedby$" in the question.

Let $k$ be a field with absolute value $|\cdot|$ and $k'$ be a complete field with absolute value $|\cdot|'$.  space. Let $i:k\to k'$ be a field embedding that is continuous. Then we have the following.
     (I). There is a unique way to extend $i$ to a continuous map from $\hat k$ to $k'$.
     (II). If $i$ is, in fact, a topological field embedding and $i(k)$ is dense in $k'$, then that extension of $i$ must be a topological field isomorphism.

