I had a troubles with this problem, so I thought it would be important to write my solution because I'm not sure of all the details.
Since I already accepted the answer, I can not open a bounty to check my solution, so I opened another question. If this is a problem, more experienced users can alert me and I will delete the question.
I tried to use Max and Arturo's hints.
Solution.
Let $(X,\varphi_{i})$ the inverse limit of $(X_{i},\varphi_{ij})$, that is, $$X = \left\{x \in \prod_{i}X_{i} \mid \varphi_{ij}\circ\pi_{j}(x) = \pi_{i}(x)\;\forall i,j; j \geq i \right\}$$ and $\varphi_{i} = \pi_{i}|_{X}$ where $\varphi_{i}: X \to X_{i}$.
We can consider $I = \mathbb{N}$
Lema 1. If $I$ is a countable direct set, then there is a cofinal countable subset totally ordered in $I$. Moreover, we can consider $\mathbb{N} \subset I$.
Proof. Since $I$ is countable, we can write $I = \{a_{j} \mid j \in \mathbb{N}\}$. Let $(x_{n})$ be a sequence defined by:
(i) $x_{0} = a_{0}$
(ii) Defined $x_{n}$, take $i = \min\{j \in \mathbb{N} \mid a_{j} \geq x_{n}; a_{j} \geq a_{n}\}$ and define $x_{n+1} = a_{i}$.
Claim: $(x_{n})$ is increasing.
We have that $x_{n+1} = a_{i}$ where $i$ satisfies $a_{i} \geq x_{n}$, that is, $x_{n+1} \geq x_{n}$.
Claim: for each $a \in I$, there is $x_{i} \geq a$.
Given $a_{k} \in I$, note that $x_{k+1} = a_{i}$ where $i$ is smallest natural that, in particular, satisfies $a_{i} \geq a_{k}$, that is, $x_{k+1} \geq a_{k}$.
We know that $\tilde{I} = \{x_{1} \leq \cdots \leq x_{n} \leq \cdots\}$ is totally ordered. Since $\tilde{I}$ is a countable subset totally ordered, $\mathbb{N}$ immerserd in $I$, that is, there is no problem to assume $\tilde{I} = \mathbb{N}$. $\Box$
Lema 2. Let $I$ be a direct set and $J \subset I$ cofinal. If $(X_{i},\varphi_{ij})$ is an inverse system indexed by $I$, then $$\varprojlim_{i \in I}X_{i} \simeq \varprojlim_{j \in J}X_{j}.$$
Proof. We have that $$\varprojlim_{i \in I}X_{i} = \left\{x \in \prod_{i}X_{i} \mid \varphi_{ik}\circ\pi_{k}(x) = \pi_{i}(x)\;\forall i,k; k \geq i \right\}$$ and $$\varprojlim_{j \in J}X_{j} = \left\{x \in \prod_{j}X_{j} \mid \varphi_{jl}\circ\pi_{l}(x) = \pi_{j}(x)\;\forall j,l; l \geq j \right\}.$$ Thus, we get a projective map $$\begin{eqnarray*} \pi: \prod_{i}X_{i} & \to & \prod_{j}X_{j}\\ x_{i} & \mapsto & x_{j}, \end{eqnarray*}$$ and, since $\pi\left(\varprojlim_{i \in I}X_{i}\right) \subset \varprojlim_{j \in J}X_{j}$, the function $$\psi = \pi|_{X}: \varprojlim_{i \in I}X_{i} \to \varprojlim_{j \in J}X_{j}$$ where $X = \varprojlim_{i \in I}X_{i}$ is well-defined.
Now, let $(x_{j}) \in \varprojlim_{j \in J}X_{j}$. If $i \in I\setminus J$ and $J$ is cofinal, there is $j \in J$ with $j \geq i$. Define $x_{i} = \varphi_{ij}(x_{j})$.
Claim. the equallity does not depends of $j$.
Let $\tilde{j} \in J$ such that $\tilde{j} \geq i$. Since $J \subset I$ and $I$ is a direct set, there is $k \in J$ such that $k \geq j$ and $k \geq \tilde{j}$. Thus, $\varphi_{jk}(x_{k}) = x_{j}$ and $\varphi_{\tilde{j}k}(x_{k}) = x_{\tilde{j}}$. Therefore, $$\begin{eqnarray*} \varphi_{ij}(x_{j}) & = & \varphi_{ij}(\varphi_{jk}(x_{k}))\\ & = & (\varphi_{ij}\circ\varphi_{jk})(x_{k})\\ & = & \varphi_{ik}(x_{k})\\ & = & (\varphi_{i\tilde{j}}\circ\varphi_{\tilde{j}k})(x_{k})\\ & = & \varphi_{i\tilde{j}}(\varphi_{\tilde{j}k}(x_{k}))\\ & = & \varphi_{i\tilde{j}}(x_{\tilde{j}}). \end{eqnarray*}$$
Given $(x_{j}) \in \varprojlim_{j \in J}X_{j}$ we get $(x_{i}) \in \prod_{i}X_{i}$ such that $\pi(x_{i}) = x_{j}$. Let $i,k \in I$ with $k \geq i$. We know that there is $j \in J$ with $j \geq k$ and so, $j \geq i$. We have that $\varphi_{kj}(x_{j}) = x_{k}$ and $\varphi_{ij}(x_{j}) = x_{i}$. Thus, $$\varphi_{ik}(\pi_{k}(x_{i})) = \varphi_{ik}(x_{k}) = \varphi_{ij}(x_{j}) = x_{i} = \pi_{i}(x_{i}),$$ that is, $(x_{i}) \in \varprojlim_{i \in I}X_{i}$. Therefore, this map is the inverse map for $\psi$. So, $\psi$ is a canonical bijection and so, the inverse limits are isomorphic. $\Box$
By lema 1 we can assume $\mathbb{N} \subset I$ is a cofinal subset of $I$ and, by lema 2, we can suppose $I = \mathbb{N}$.
Now, take $x_{0} \in X_{0}$ and, since $X_{i} \neq \emptyset$ and $\varphi_{ij}$ is surjective, define indutivelly $x_{n} \in E_{n}$ so that $x_{n} \in \varphi^{-1}_{n-1,n}(x_{n-1})$ for $n \geq 1$.
Claim. $x_{m} = \varphi_{m,n}(x_{n})$ for $m \leq n$.
If $n-m = 0$, then $\varphi_{nn}(x_{n}) = x_{n}$ by hypothesis. If $n-m=1$, then by definition of $x_{n}$, $\varphi_{n-1,n}(x_{n}) = x_{n-1}$. If the result is true for $n-m = k$, then $$\varphi_{n-k-1,n}(x_{n}) = \varphi_{n-k-1,n-k}\circ\varphi_{n-k,n}(x_{n}) = \varphi_{n-k-1,n-k}(x_{n-k}) = x_{n-k-1}.$$ Now, whereas $\varphi_{i} = \pi_{i}|_{X}: X \to X_{i}$, we have that $$\varphi_{mn}\circ\pi_{n}(x_{n}) = \varphi_{mn}(x_{n}) = x_{m} = \pi_{m}(x_{n}),$$ therefore, $(x_{n}) \in X$.
Your proof is fine. Let me advertise for universal properties though, to show how the claim that you can take a cofinal $J$ is proved very smoothly with them.
So with the same notations, $J\subset I$ is cofinal. Then let $X:= \varprojlim_{J} X_i$ . For $i\in I$, note that whichever $ j\geq i$ I choose, the projection map to $j$ followed by $X_j \to X_i$ is the same. So I have projections to each $X_i, i\in I$ (because $J$ is cofinal) which are of course compatible with all the maps $X_k\to X_l$.
Let $A$ be any object of your category, and suppose you have compatible maps $ A\to X_i$ for $i\in I$. Then you have them for $i\in J$ and they're still compatible, so you get a unique map $A\to X$ which commutes with the $J$ projections. But now (draw a digaram) by definition of the $I$ projections of $X$, and because the maps $A\to X_i$ are compatible, this easily implies that the map $A\to X$ is compatible with all the $I$ projections.
Therefore $X$ has the universal property of the inverse limit, thus it is isomorphic to it.